All Exams >
NEET >
NCERT Based Tests for NEET >
All Questions
All questions of Electric Charges and Fields for NEET Exam
Two point charges of +3 μC and +4 μC repel each other with a force of 10 N. If each is given an additional charge of -6 μC, the new force is
- a)2N
- b)4N
- c)7.5N
- d)5N
Correct answer is option 'D'. Can you explain this answer?
Two point charges of +3 μC and +4 μC repel each other with a force of 10 N. If each is given an additional charge of -6 μC, the new force is
a)
2N
b)
4N
c)
7.5N
d)
5N
|
Raghav Bansal answered |
The force between two charges q1 and q2 at a distance r is given by Coulomb's law:
F = k*q1*q2/r^2
where k is Coulomb's constant (9*10^9 N*m^2/C^2).
Given that F = 10 N, q1 = 3 μC = 3*10^-6 C, and q2 = 4 μC = 4*10^-6 C, we can solve for r^2:
10 = 9*10^9 * 3*10^-6 * 4*10^-6 / r^2
r^2 = 9*10^9 * 3*10^-6 * 4*10^-6 / 10
r^2 = 1.08*10^-2 m^2
Now, if each charge is given an additional -6 μC, the new charges are q1' = -3 μC = -3*10^-6 C and q2' = -2 μC = -2*10^-6 C. The new force F' is:
F' = k*q1'*q2'/r^2
F' = 9*10^9 * -3*10^-6 * -2*10^-6 / 1.08*10^-2
F' = 5 N
So, the new force is 5N. Hence, option
F = k*q1*q2/r^2
where k is Coulomb's constant (9*10^9 N*m^2/C^2).
Given that F = 10 N, q1 = 3 μC = 3*10^-6 C, and q2 = 4 μC = 4*10^-6 C, we can solve for r^2:
10 = 9*10^9 * 3*10^-6 * 4*10^-6 / r^2
r^2 = 9*10^9 * 3*10^-6 * 4*10^-6 / 10
r^2 = 1.08*10^-2 m^2
Now, if each charge is given an additional -6 μC, the new charges are q1' = -3 μC = -3*10^-6 C and q2' = -2 μC = -2*10^-6 C. The new force F' is:
F' = k*q1'*q2'/r^2
F' = 9*10^9 * -3*10^-6 * -2*10^-6 / 1.08*10^-2
F' = 5 N
So, the new force is 5N. Hence, option
If 109 electrons move out of a body to another body every second, then the time required to get charge of 1 C on the other body is - a)250 years
- b)100 years
- c)198 years
- d)150 years
Correct answer is option 'C'. Can you explain this answer?
If 109 electrons move out of a body to another body every second, then the time required to get charge of 1 C on the other body is
a)
250 years
b)
100 years
c)
198 years
d)
150 years
|
Priya Menon answered |
The charge given out in one second = 1.6 x 10-19 C x 109 = 1.6 x 10-10C
Time required to accumulate a charge of 1 C
= 1/(1.6 x 10-10) = 6.25 x 109 s = 198 years
Time required to accumulate a charge of 1 C
= 1/(1.6 x 10-10) = 6.25 x 109 s = 198 years
The electric field at a point is- a)always continuous
- b)continuous if there is no charge at that point
- c)discontinuous if there is a charge at that point
- d)both (b) and (c) are correct
Correct answer is option 'D'. Can you explain this answer?
The electric field at a point is
a)
always continuous
b)
continuous if there is no charge at that point
c)
discontinuous if there is a charge at that point
d)
both (b) and (c) are correct
|
Meera Singh answered |
Electric field at a point is continuous if there is no charge at that point. And the field is discontinuous if there is charge at that point. So both options (b) and (c) are correct.
Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in figure. The force on a charge of 1μC placed at the centre of square is
- a)7N
- b)8N
- c)2N
- d)zero
Correct answer is option 'D'. Can you explain this answer?
Four point charges are placed at the corners of a square ABCD of side 10 cm, as shown in figure. The force on a charge of 1μC placed at the centre of square is
a)
7N
b)
8N
c)
2N
d)
zero
|
Suresh Iyer answered |
Forces of repulsion on 1μC charge at O due to 3 μC charge, at A and C are equal and opposite. So they cancel each other.
Similarly, forces of attraction of 1μC charge at O due to -4μC charges at B and D are also equal and opposite. So they also cancel each other.
Hence the net force on the charge of 1μC at O is zero.
Similarly, forces of attraction of 1μC charge at O due to -4μC charges at B and D are also equal and opposite. So they also cancel each other.
Hence the net force on the charge of 1μC at O is zero.
Which of the following curves shown below cannot possibly represent electrostatic field lines?
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which of the following curves shown below cannot possibly represent electrostatic field lines?
a)
b)
c)
d)
|
|
Riya Banerjee answered |
Perpendicular to the Surface of Conductors: Electrostatic field lines are always perpendicular (at 90 degrees) to the surface of a conductor. This is because conductors contain free charges that can move freely within the material. When an electric field is applied, these free charges redistribute themselves on the surface of the conductor until the electric field inside the conductor is zero. This redistribution causes the electric field at the surface to be perpendicular to the surface.
Two parallel infinite line charges + λ and -λ are placed with a separation distance R in free space. Then etelectric field exactly mid - way between the two line charges is - a) zero
- b)2λ/πε0R
- c)λ/πε0R
- d)λ/2πε0R
Correct answer is option 'B'. Can you explain this answer?
Two parallel infinite line charges + λ and -λ are placed with a separation distance R in free space. Then etelectric field exactly mid - way between the two line charges is
a)
zero
b)
2λ/πε0R
c)
λ/πε0R
d)
λ/2πε0R
|
|
Raghav Bansal answered |
Electric field at point P due to line charge distribution +λ,
E+λ =
away from +λ
Electric field at point P due to line charge distribution -λ,
E+λ and E-λ have same direction,
E+λ =
away from +λElectric field at point P due to line charge distribution -λ,
E+λ and E-λ have same direction,
A polythene piece rubbed with wool is found to have a negative charge of 6 x 10-7 C. The number of electrons transferred to polythene from wool is - a)3.75 x 1010
- b)9.6 x 1010
- c)9.6 x 1012
- d)3.75 x 1012
Correct answer is option 'D'. Can you explain this answer?
A polythene piece rubbed with wool is found to have a negative charge of 6 x 10-7 C. The number of electrons transferred to polythene from wool is
a)
3.75 x 1010
b)
9.6 x 1010
c)
9.6 x 1012
d)
3.75 x 1012
|
Dev Patel answered |
Here, q = -6 x 10-7 C
∴ Number of electrons transferred from wool to polythene, n =
∴ Number of electrons transferred from wool to polythene, n =
An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2 x 104N/C. The time taken by the electron to fall this distance is - a)1.3 x 102 s
- b)2.1 x 10-12 s
- c)1.6 x 10-10 s
- d)2.9 x 10-9 s
Correct answer is option 'D'. Can you explain this answer?
An electron initially at rest falls a distance of 1.5 cm in a uniform electric field of magnitude 2 x 104N/C. The time taken by the electron to fall this distance is
a)
1.3 x 102 s
b)
2.1 x 10-12 s
c)
1.6 x 10-10 s
d)
2.9 x 10-9 s
|
|
Mira Joshi answered |
In figure the field is upward. So the negatively charged electron experiences a downward force.
∴ The acceleration of electron is
ac = eE/me
The time required by the electron to fall through a distance h is
∴ The acceleration of electron is
ac = eE/me
The time required by the electron to fall through a distance h is
Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell is- a)inversely proportional to σ
- b)directly proportional to x2
- c)directly proportional to R
- d)inversely proportional to x2
Correct answer is option 'D'. Can you explain this answer?
Consider a thin spherical shell of radius R consisting of uniform surface charge density σ. The electric field at a point of distance x from its centre and outside the shell is
a)
inversely proportional to σ
b)
directly proportional to x2
c)
directly proportional to R
d)
inversely proportional to x2
|
|
Mira Joshi answered |
For a thin uniformly charged spherical shell, the field points outside the shell at a distance x from the centre is
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
If the radius of the sphere is R, Q = σ4πR2
This is inversely proportional to square of the distance from the centre. It is as if the whole charge is concentrated at the centre.
Electrical as well as gravitational affects acan be thought to be caused by fields. Which the following is true of an electrical or gravitational field?- a)The field concept is often used to describe contact forces.
- b)Gravitational or electric field does not exit in the space around an object.
- c)Fields are useful for understanding forces acting through a distance.
- d)There is no way to verify the existence of a force field since it is just a concept.
Correct answer is option 'C'. Can you explain this answer?
Electrical as well as gravitational affects acan be thought to be caused by fields. Which the following is true of an electrical or gravitational field?
a)
The field concept is often used to describe contact forces.
b)
Gravitational or electric field does not exit in the space around an object.
c)
Fields are useful for understanding forces acting through a distance.
d)
There is no way to verify the existence of a force field since it is just a concept.
|
|
Suresh Iyer answered |
In physics, concept of field is a model used to explain the influence that a massive body or charged particle extends into the space around itself, producing a force on another massive body or charged body placed in space. Thus, concept of field is used to explain gravitational and electrostatic phenomena.
Electric field lines provide information about- a)field strength
- b)direction
- c)nature of charge
- d)all of these
Correct answer is option 'D'. Can you explain this answer?
Electric field lines provide information about
a)
field strength
b)
direction
c)
nature of charge
d)
all of these
|
Neha Dasgupta answered |
Electric field lines provide information about the field strength, direction, and nature of charge. Let's break down each aspect of this answer.
Field Strength:
Electric field lines are closer together where the field is stronger and farther apart where it is weaker. Thus, the density of field lines gives an idea about the strength of the electric field at different points. For example, if the field lines are closer together near a charged object, it means that the electric field is stronger at that point.
Direction:
The direction of the electric field lines gives information about the direction of the electric field at different points. The direction of the field lines is from positive to negative charges. Thus, the electric field lines always point away from positive charges and towards negative charges.
Nature of Charge:
Electric field lines give information about the nature of charge. If the field lines originate from a positive charge, it means that the charge is positive. Similarly, if the field lines end at a negative charge, it means that the charge is negative.
Conclusion:
In summary, electric field lines provide comprehensive information about the electric field. By analyzing the density, direction, and nature of charges, one can understand a lot about the electric field. Therefore, the correct answer is option 'D' - all of these.
Field Strength:
Electric field lines are closer together where the field is stronger and farther apart where it is weaker. Thus, the density of field lines gives an idea about the strength of the electric field at different points. For example, if the field lines are closer together near a charged object, it means that the electric field is stronger at that point.
Direction:
The direction of the electric field lines gives information about the direction of the electric field at different points. The direction of the field lines is from positive to negative charges. Thus, the electric field lines always point away from positive charges and towards negative charges.
Nature of Charge:
Electric field lines give information about the nature of charge. If the field lines originate from a positive charge, it means that the charge is positive. Similarly, if the field lines end at a negative charge, it means that the charge is negative.
Conclusion:
In summary, electric field lines provide comprehensive information about the electric field. By analyzing the density, direction, and nature of charges, one can understand a lot about the electric field. Therefore, the correct answer is option 'D' - all of these.
In the previous question , what will be the electric field at O if the charge q at A is replaced by -q?
- a)(q/4πε0r2) along OB
- b)(2q/4πε0r2) along OA
- c)(4q2/4πε0r2) along OC
- d)zero
Correct answer is option 'B'. Can you explain this answer?
In the previous question , what will be the electric field at O if the charge q at A is replaced by -q?
a)
(q/4πε0r2) along OB
b)
(2q/4πε0r2) along OA
c)
(4q2/4πε0r2) along OC
d)
zero
|
|
Suresh Iyer answered |
If the charge q at A is replaced by -q, it is equivalent to adding charge -2q at A. Therefore, electric field at O, E2 = (2q/4πε0r2) along OA.
The unit of electric dipole moment is- a)newton
- b)coulomb
- c)farad
- d)debye
Correct answer is option 'D'. Can you explain this answer?
The unit of electric dipole moment is
a)
newton
b)
coulomb
c)
farad
d)
debye
|
|
Mira Joshi answered |
The unit of electric dipole moment is: 2ql = C − m = Debye
Two point charges of 1μC and -1μC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be - a)9 N C-1
- b)0.9 N C-1
- c)90 N C-1
- d)0.09 N C-1
Correct answer is option 'D'. Can you explain this answer?
Two point charges of 1μC and -1μC are separated by a distance of 100 Å. A point P is at a distance of 10 cm from the midpoint and on the perpendicular bisector of the line joining the two charges. The electric field at P will be
a)
9 N C-1
b)
0.9 N C-1
c)
90 N C-1
d)
0.09 N C-1
|
|
Priya Menon answered |
The Point lies on equatorial line of a short dipole,
A uniformly charged conducting sphere of 4.4m diameter has a surface charge density of 60 μC m-2. The charge on the sphere is- a)7.3 x 10-3C
- b)3.7 x 10-6C
- c)7.3 x 10-6 C
- d)3.7 x 10-3 C
Correct answer is option 'D'. Can you explain this answer?
A uniformly charged conducting sphere of 4.4m diameter has a surface charge density of 60 μC m-2. The charge on the sphere is
a)
7.3 x 10-3C
b)
3.7 x 10-6C
c)
7.3 x 10-6 C
d)
3.7 x 10-3 C
|
Anjali Sharma answered |
Here D = 2r = 4.4 m,
or r = 2.2 m, σ = 60 μC m-2
Charge on the sphere, q = σ x 4πr2
= 60 x 10-6 x 4 x (22/7) x (2.2)2 = 3.7 x 10-3 C.
or r = 2.2 m, σ = 60 μC m-2
Charge on the sphere, q = σ x 4πr2
= 60 x 10-6 x 4 x (22/7) x (2.2)2 = 3.7 x 10-3 C.
The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.4 C is- a)2.5 x 1019
- b)1.5 x 1019
- c)1.5 x 10-19
- d)2.5 x 10-19
Correct answer is option 'B'. Can you explain this answer?
The number of electrons that must be removed from an electrically neutral silver dollar to give it a charge of +2.4 C is
a)
2.5 x 1019
b)
1.5 x 1019
c)
1.5 x 10-19
d)
2.5 x 10-19
|
Vivek Patel answered |
Total charge, q = 2.4 C
Then by quantization of charge, q = ne
∴ number of electrons, n = q/e = 2.4C/1.6 x 10-19 C = 1.5 x 1019
Then by quantization of charge, q = ne
∴ number of electrons, n = q/e = 2.4C/1.6 x 10-19 C = 1.5 x 1019
The force per unit charge is known as
- a)electric flux
- b)electric field
- c)electric potential
- d)electric current
Correct answer is option 'B'. Can you explain this answer?
The force per unit charge is known as
a)
electric flux
b)
electric field
c)
electric potential
d)
electric current
|
|
Anjali Sharma answered |
The force per unit charge is known as: E= Fq
This formula f is hold for electric field.
This formula f is hold for electric field.
The electrostatic attracting force on a small sphere of charge 0.2μC due to another small sphere of charge -0.4μC in air is 0.4N. The distance between the two spheres is- a)43.2 x 10-6 m
- b)42.4 x 10-3 m
- c)18.1 x 10-3 m
- d)19.2 x 10-6 m
Correct answer is option 'B'. Can you explain this answer?
The electrostatic attracting force on a small sphere of charge 0.2μC due to another small sphere of charge -0.4μC in air is 0.4N. The distance between the two spheres is
a)
43.2 x 10-6 m
b)
42.4 x 10-3 m
c)
18.1 x 10-3 m
d)
19.2 x 10-6 m
|
|
Mira Joshi answered |
Here, q1 = 0.2 μC = 0.2 x 10-6 C
q2 = 0.4 μC = -0.4 x 10-6 C, F = -0.4 N
As F =
= 1.8 x 10-3
∴ r = (1.8 x 10-3)1/2 = 0.0424 m = 42.4 x 10-3 m
q2 = 0.4 μC = -0.4 x 10-6 C, F = -0.4 N
As F =
= 1.8 x 10-3
∴ r = (1.8 x 10-3)1/2 = 0.0424 m = 42.4 x 10-3 m
The constant k in Coulomb's law depends on- a)nature of medium
- b)system of units
- c)intensity of charge
- d)both (a) and (b)
Correct answer is option 'A'. Can you explain this answer?
The constant k in Coulomb's law depends on
a)
nature of medium
b)
system of units
c)
intensity of charge
d)
both (a) and (b)
|
Gaurav Kumar answered |
The value of k = (1/4πε0) = 8.854 x 10-12 C2 N-1 m-2 where ε0 is permittivity of free space.
When a person combs his hair, static electricity is sometimes generated by what process?- a)Contact between the comb and hair results in a charge.
- b)Friction between the comb and hair results in the transfer of electrons.
- c)Deduction between the comb and hair.
- d)Induction between the comb hair.
Correct answer is option 'B'. Can you explain this answer?
When a person combs his hair, static electricity is sometimes generated by what process?
a)
Contact between the comb and hair results in a charge.
b)
Friction between the comb and hair results in the transfer of electrons.
c)
Deduction between the comb and hair.
d)
Induction between the comb hair.
|
|
Yashvi Malik answered |
Contact and Friction
When a person combs their hair, static electricity is sometimes generated. This is because of the process of contact and friction between the comb and the hair.
Transfer of Electrons
During the combing process, the comb and the hair come into contact with each other. This contact leads to the transfer of electrons between the two surfaces.
Electrostatic Charge
Electrons, which have a negative charge, are loosely bound to the atoms in the hair. When the comb is run through the hair, the friction between the comb and the hair causes some of these electrons to be transferred from the hair to the comb.
Positive and Negative Charges
As a result of this transfer of electrons, the comb becomes negatively charged, while the hair becomes positively charged. This creates an imbalance of charges and generates static electricity.
Electrostatic Discharge
When the negatively charged comb is removed from the hair, it can attract nearby positively charged objects or cause tiny hairs to stand up due to repulsion between like charges. This is often observed as hair standing on end or a small shock when touching a metal object.
Explanation of the Correct Answer
The correct answer to this question is option B: Friction between the comb and hair results in the transfer of electrons. This is because the process of combing involves contact and friction, which leads to the transfer of electrons between the comb and the hair, generating static electricity.
When a person combs their hair, static electricity is sometimes generated. This is because of the process of contact and friction between the comb and the hair.
Transfer of Electrons
During the combing process, the comb and the hair come into contact with each other. This contact leads to the transfer of electrons between the two surfaces.
Electrostatic Charge
Electrons, which have a negative charge, are loosely bound to the atoms in the hair. When the comb is run through the hair, the friction between the comb and the hair causes some of these electrons to be transferred from the hair to the comb.
Positive and Negative Charges
As a result of this transfer of electrons, the comb becomes negatively charged, while the hair becomes positively charged. This creates an imbalance of charges and generates static electricity.
Electrostatic Discharge
When the negatively charged comb is removed from the hair, it can attract nearby positively charged objects or cause tiny hairs to stand up due to repulsion between like charges. This is often observed as hair standing on end or a small shock when touching a metal object.
Explanation of the Correct Answer
The correct answer to this question is option B: Friction between the comb and hair results in the transfer of electrons. This is because the process of combing involves contact and friction, which leads to the transfer of electrons between the comb and the hair, generating static electricity.
Two insulated charged metallic spheres P and Q have their centres separated by a distance of 60 cm. The radii of P and Q are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2 x 10-7 C is- a)5.2 x 10-4 N
- b)2.5 x 10-3 N
- c)1.5 x 10-3 N
- d)3.5 x 10-4 N
Correct answer is option 'B'. Can you explain this answer?
Two insulated charged metallic spheres P and Q have their centres separated by a distance of 60 cm. The radii of P and Q are negligible compared to the distance of separation. The mutual force of electrostatic repulsion if the charge on each is 3.2 x 10-7 C is
a)
5.2 x 10-4 N
b)
2.5 x 10-3 N
c)
1.5 x 10-3 N
d)
3.5 x 10-4 N
|
|
Riya Banerjee answered |
Here, q1 = q2 = 3.2 x 10-7 C, r = 60cm = 0.6 m
Electrostatic force,
Electrostatic force,
In the above question , the flux through the same square if the line normal to its plane makes a 60° angle with the x-axis is
- a)30 N C-1 m2
- b)10 N C-1 m2
- c)20 N C-1 m2
- d)25 N C-1 m2
Correct answer is option 'B'. Can you explain this answer?
In the above question , the flux through the same square if the line normal to its plane makes a 60° angle with the x-axis is
a)
30 N C-1 m2
b)
10 N C-1 m2
c)
20 N C-1 m2
d)
25 N C-1 m2
|
|
Anjali Sharma answered |
When normal to the plane makes an angle of 60° with x-axis, then θ = 60°
φ = Es cosθ = 2 x 103 x 10-2 cos 60° = 10 N C-1m2
φ = Es cosθ = 2 x 103 x 10-2 cos 60° = 10 N C-1m2
An electric dipole consists of charges ±2.0 x 10-8C separated by a distance of 2.0 x 10-3 m. It is placed near a long line charge of linear charge density 4.0 x 10-4C m-1 as shown in the figure, such that the negative charge is at a distance of 2.0 cm from the line charge. The force acting on the dipole will be- a)7.2 N towards the line charge
- b)6.6 N away from the line charge
- c)0.6 N away from the line charge
- d)0.6 N towards the line charge
Correct answer is option 'D'. Can you explain this answer?
An electric dipole consists of charges ±2.0 x 10-8C separated by a distance of 2.0 x 10-3 m. It is placed near a long line charge of linear charge density 4.0 x 10-4C m-1 as shown in the figure, such that the negative charge is at a distance of 2.0 cm from the line charge. The force acting on the dipole will be
a)
7.2 N towards the line charge
b)
6.6 N away from the line charge
c)
0.6 N away from the line charge
d)
0.6 N towards the line charge
|
Hansa Sharma answered |
The electric field at a distance r from the line charge of linear density λ is given by
E = λ/2πε0r
Hence, the field at the negative charge,
E1 =
The force on the negative charge, F1 = (3.6 x 108) (2.0 x 10-8) = 7.2 N towards the line charge
Similarly, the field at the positive charge,
i.e., at r = 0.022 m is E2 = 3.3 x108 N -1
The force on the positive charge,
F2 = (3.3 x 108) x (2.0 x 10-8) = 6.6 N away from the line charge.
Hence, the net force on the dipole = 7.2N - 6.6N = 0.6 N towards the line charge.
E = λ/2πε0r
Hence, the field at the negative charge,
E1 =
The force on the negative charge, F1 = (3.6 x 108) (2.0 x 10-8) = 7.2 N towards the line charge
Similarly, the field at the positive charge,
i.e., at r = 0.022 m is E2 = 3.3 x108 N -1
The force on the positive charge,
F2 = (3.3 x 108) x (2.0 x 10-8) = 6.6 N away from the line charge.
Hence, the net force on the dipole = 7.2N - 6.6N = 0.6 N towards the line charge.
A cup contains 250 g of water. Find the total positive charges present in the cup of water.- a)1.34 x 1019 C
- b)1.34 x 107 C
- c)2.43 x 1019 C
- d)2.43 x 107 C
Correct answer is option 'B'. Can you explain this answer?
A cup contains 250 g of water. Find the total positive charges present in the cup of water.
a)
1.34 x 1019 C
b)
1.34 x 107 C
c)
2.43 x 1019 C
d)
2.43 x 107 C
|
|
Geetika Shah answered |
Mass of water = 250 g,
Molecular mass of water = 18 g
Number of molecules in 18 g of water = 6.02 x 1023
Number of molecules in one cup of water = (250/18) x 6.02 x 1023
Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.
∴ Total positive charge present in one cup of water
= (250/18) x 6.02 x 1023 x 10 x 1.6 x 10-19 C = 1.34 x 107C.
Molecular mass of water = 18 g
Number of molecules in 18 g of water = 6.02 x 1023
Number of molecules in one cup of water = (250/18) x 6.02 x 1023
Each molecule of water contains two hydrogen atoms and one oxygen atom, i.e., 10 electrons and 10 protons.
∴ Total positive charge present in one cup of water
= (250/18) x 6.02 x 1023 x 10 x 1.6 x 10-19 C = 1.34 x 107C.
Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region- a)the electric field is necessarily zero.
- b)the electric field is due to the dipole moment of the charge distribution only.
- c)the dominant electric field is inversely proportional to r3, for large r (distance from origin).
- d)the work done to move a charged particle along a closed path, away from the region will not be zero.
Correct answer is option 'C'. Can you explain this answer?
Consider a region inside which, there are various types of charges but the total charge is zero. At points outside the region
a)
the electric field is necessarily zero.
b)
the electric field is due to the dipole moment of the charge distribution only.
c)
the dominant electric field is inversely proportional to r3, for large r (distance from origin).
d)
the work done to move a charged particle along a closed path, away from the region will not be zero.
|
|
Riya Banerjee answered |
When there are various types of charges in a region, but the total charge is zero, the region can be supposed to contain a number of electric dipoles. Therefore, at points outside the region, the dominant electric field ∝ (1/r3) for large r.
Under the action of a given coulombic force the acceleration of an electron is 2.5 x 1022 ms-2. Then the magnitude of the acceleration of a proton under the action of same force is nearly- a)1.6 x 10-19 ms-2
- b)9.1 x 1031 ms-2
- c)1.5 x 1019 ms-2
- d)1.6 x 1027 ms-2
Correct answer is option 'C'. Can you explain this answer?
Under the action of a given coulombic force the acceleration of an electron is 2.5 x 1022 ms-2. Then the magnitude of the acceleration of a proton under the action of same force is nearly
a)
1.6 x 10-19 ms-2
b)
9.1 x 1031 ms-2
c)
1.5 x 1019 ms-2
d)
1.6 x 1027 ms-2
|
|
Kavya Sarkar answered |
Given:
- Acceleration of an electron under the action of a coulombic force = 2.5 x 10^22 m/s^2
To find:
Magnitude of the acceleration of a proton under the action of the same force.
Explanation:
The coulombic force experienced by a charged particle is given by Coulomb's law:
F = k * (q1 * q2) / r^2
Where,
F = Force between two charged particles
k = Coulomb's constant (9 x 10^9 Nm^2/C^2)
q1, q2 = Charges of the particles
r = Distance between the particles
The acceleration of a charged particle is given by Newton's second law of motion:
F = m * a
Where,
m = Mass of the particle
a = Acceleration of the particle
Comparing the forces on the electron and the proton:
Since the force is the same for both the electron and the proton, we can equate the coulombic forces on them:
k * (q_e * q_p) / r^2 = k * (q_e * q_e) / r^2
Key point 1:
The charge of an electron (q_e) is -1.6 x 10^-19 C, and the charge of a proton (q_p) is +1.6 x 10^-19 C. Therefore,
-1.6 x 10^-19 * q_p = -1.6 x 10^-19 * -1.6 x 10^-19
Key point 2:
The mass of an electron (m_e) is 9.1 x 10^-31 kg, and the mass of a proton (m_p) is 1.67 x 10^-27 kg.
Substituting the values into the equation:
Using the equation F = m * a, we can solve for the acceleration of the proton:
k * (q_e * q_p) / r^2 = m_p * a_p
Key point 3:
We know that the acceleration of the electron (a_e) is 2.5 x 10^22 m/s^2.
k * (q_e * q_e) / r^2 = m_e * a_e
Solving for a_p:
We can rearrange the equation to solve for the acceleration of the proton:
a_p = (k * (q_e * q_p) / r^2) / m_p
Substituting the values:
Plugging in the given values, we can calculate the acceleration of the proton:
a_p = (9 x 10^9 * (-1.6 x 10^-19 * 1.6 x 10^-19) / r^2) / (1.67 x 10^-27)
Key point 4:
The radius of the orbit (r) is not given, so we cannot calculate the exact value of the acceleration of the proton. However, we can compare it to the acceleration of the electron.
Comparing the accelerations:
Since the magnitude of the charge of the proton is the same as the magnitude of the charge of
- Acceleration of an electron under the action of a coulombic force = 2.5 x 10^22 m/s^2
To find:
Magnitude of the acceleration of a proton under the action of the same force.
Explanation:
The coulombic force experienced by a charged particle is given by Coulomb's law:
F = k * (q1 * q2) / r^2
Where,
F = Force between two charged particles
k = Coulomb's constant (9 x 10^9 Nm^2/C^2)
q1, q2 = Charges of the particles
r = Distance between the particles
The acceleration of a charged particle is given by Newton's second law of motion:
F = m * a
Where,
m = Mass of the particle
a = Acceleration of the particle
Comparing the forces on the electron and the proton:
Since the force is the same for both the electron and the proton, we can equate the coulombic forces on them:
k * (q_e * q_p) / r^2 = k * (q_e * q_e) / r^2
Key point 1:
The charge of an electron (q_e) is -1.6 x 10^-19 C, and the charge of a proton (q_p) is +1.6 x 10^-19 C. Therefore,
-1.6 x 10^-19 * q_p = -1.6 x 10^-19 * -1.6 x 10^-19
Key point 2:
The mass of an electron (m_e) is 9.1 x 10^-31 kg, and the mass of a proton (m_p) is 1.67 x 10^-27 kg.
Substituting the values into the equation:
Using the equation F = m * a, we can solve for the acceleration of the proton:
k * (q_e * q_p) / r^2 = m_p * a_p
Key point 3:
We know that the acceleration of the electron (a_e) is 2.5 x 10^22 m/s^2.
k * (q_e * q_e) / r^2 = m_e * a_e
Solving for a_p:
We can rearrange the equation to solve for the acceleration of the proton:
a_p = (k * (q_e * q_p) / r^2) / m_p
Substituting the values:
Plugging in the given values, we can calculate the acceleration of the proton:
a_p = (9 x 10^9 * (-1.6 x 10^-19 * 1.6 x 10^-19) / r^2) / (1.67 x 10^-27)
Key point 4:
The radius of the orbit (r) is not given, so we cannot calculate the exact value of the acceleration of the proton. However, we can compare it to the acceleration of the electron.
Comparing the accelerations:
Since the magnitude of the charge of the proton is the same as the magnitude of the charge of
A conducting sphere of radius 10 cm has unknown charge. If the electric field at a distance 20 cm from the centre of the sphere is 1.2 x 103N C-1 and points radially inwards. The net charge on the sphere is- a)-4.5 x 10-9 C
- b)4.5 x 109 C
- c)-5.3 x 10-9 C
- d)5.3 x 109 C
Correct answer is option 'C'. Can you explain this answer?
A conducting sphere of radius 10 cm has unknown charge. If the electric field at a distance 20 cm from the centre of the sphere is 1.2 x 103N C-1 and points radially inwards. The net charge on the sphere is
a)
-4.5 x 10-9 C
b)
4.5 x 109 C
c)
-5.3 x 10-9 C
d)
5.3 x 109 C
|
Nishtha Joshi answered |
Understanding the Problem
A conducting sphere with a radius of 10 cm has an unknown charge. The electric field at a distance of 20 cm from the center of the sphere is given as 1.2 x 10^3 N/C and points radially inwards, indicating that the sphere has a net negative charge.
Key Concepts
- Electric Field Due to a Charged Sphere: For a conducting sphere, the electric field outside the sphere behaves as though all the charge were concentrated at the center. The formula for the electric field (E) at a distance (r) from the center is given by:
E = k * |Q| / r²
where k is the Coulomb's constant (approximately 8.99 x 10^9 Nm²/C²) and Q is the net charge.
- Direction of the Electric Field: The inward direction of the electric field indicates that the charge on the sphere is negative.
Calculating the Charge
Given:
- E = 1.2 x 10^3 N/C
- r = 20 cm = 0.2 m
- k = 8.99 x 10^9 Nm²/C²
Substituting the values into the electric field equation:
1. Rearranging gives: Q = E * r² / k
2. Plugging the values:
Q = (1.2 x 10^3 N/C) * (0.2 m)² / (8.99 x 10^9 Nm²/C²)
3. Calculating:
Q = (1.2 x 10^3) * (0.04) / (8.99 x 10^9)
Q = -5.3 x 10^-9 C (the negative sign shows the charge is negative)
Conclusion
The net charge on the conducting sphere is approximately -5.3 x 10^-9 C, confirming that option 'C' is correct.
A conducting sphere with a radius of 10 cm has an unknown charge. The electric field at a distance of 20 cm from the center of the sphere is given as 1.2 x 10^3 N/C and points radially inwards, indicating that the sphere has a net negative charge.
Key Concepts
- Electric Field Due to a Charged Sphere: For a conducting sphere, the electric field outside the sphere behaves as though all the charge were concentrated at the center. The formula for the electric field (E) at a distance (r) from the center is given by:
E = k * |Q| / r²
where k is the Coulomb's constant (approximately 8.99 x 10^9 Nm²/C²) and Q is the net charge.
- Direction of the Electric Field: The inward direction of the electric field indicates that the charge on the sphere is negative.
Calculating the Charge
Given:
- E = 1.2 x 10^3 N/C
- r = 20 cm = 0.2 m
- k = 8.99 x 10^9 Nm²/C²
Substituting the values into the electric field equation:
1. Rearranging gives: Q = E * r² / k
2. Plugging the values:
Q = (1.2 x 10^3 N/C) * (0.2 m)² / (8.99 x 10^9 Nm²/C²)
3. Calculating:
Q = (1.2 x 10^3) * (0.04) / (8.99 x 10^9)
Q = -5.3 x 10^-9 C (the negative sign shows the charge is negative)
Conclusion
The net charge on the conducting sphere is approximately -5.3 x 10^-9 C, confirming that option 'C' is correct.
A uniform electric field E = 2 x 103 N C-1 is acting along the positive x-axis. The flux of this field through a square of 10cm on a side whose plane is paralled to the yz plane is - a)20 N C-1 m2
- b)30 N C-1 m2
- c)10 N C-1 m2
- d)40 N C-1 m2
Correct answer is option 'A'. Can you explain this answer?
A uniform electric field E = 2 x 103 N C-1 is acting along the positive x-axis. The flux of this field through a square of 10cm on a side whose plane is paralled to the yz plane is
a)
20 N C-1 m2
b)
30 N C-1 m2
c)
10 N C-1 m2
d)
40 N C-1 m2
|
|
Riya Banerjee answered |
Here, E = 2 x 103 N C-1 is along + x-axis
Surface area, s = (10 cm)2 = 102 x 10-4 m2 = 10-2 m2
When plane is parallel to yz plane, θ = 0°
So φ = E s cosθ = 2 x 103 x 10-2 cos 0° = 20N C-1 m2
Surface area, s = (10 cm)2 = 102 x 10-4 m2 = 10-2 m2
When plane is parallel to yz plane, θ = 0°
So φ = E s cosθ = 2 x 103 x 10-2 cos 0° = 20N C-1 m2
The surface considered for Gauss's law is called- a)Closed surface
- b)Spherical surface
- c)Gaussian surface
- d)Plane surface
Correct answer is option 'C'. Can you explain this answer?
The surface considered for Gauss's law is called
a)
Closed surface
b)
Spherical surface
c)
Gaussian surface
d)
Plane surface
|
|
Riya Banerjee answered |
The surface that we choose for the application of Gauss's law is called Gaussian surface.
A circular plane sheet of radius 10 cm is placed in a uniform electric field of 5 x 105N C-1, making an angle of 600 with the field. The electric flux through the sheet is- a)1.36 x 102 N m2 C-1
- b)1.36 x 104 N m2 C-1
- c)0.515 x 102 N m2 C-1
- d)0.515 x 104 N m2 C-1
Correct answer is option 'B'. Can you explain this answer?
A circular plane sheet of radius 10 cm is placed in a uniform electric field of 5 x 105N C-1, making an angle of 600 with the field. The electric flux through the sheet is
a)
1.36 x 102 N m2 C-1
b)
1.36 x 104 N m2 C-1
c)
0.515 x 102 N m2 C-1
d)
0.515 x 104 N m2 C-1
|
|
Anjali Sharma answered |
Here, r = 10 cm 0.1 m, E = 5 x 105 N C-1
As the angle between the plane sheet and the electric field is 600, angle made by the normal to the plane sheet and the electric field is θ = 900 - 600 = 300
φE = ES cos θ = E x πr2 cos θ
= 5 x 105 x 3.14 x (0.1)2 cos 300 = 1.36 x 104 N m2C-1
As the angle between the plane sheet and the electric field is 600, angle made by the normal to the plane sheet and the electric field is θ = 900 - 600 = 300
φE = ES cos θ = E x πr2 cos θ
= 5 x 105 x 3.14 x (0.1)2 cos 300 = 1.36 x 104 N m2C-1
Which of the following statements about dipole moment is not true?- a)The dimensions of dipole moment is [LTA].
- b)The unit of dipole moment of C m.
- c)Dipole moment is vector quantity and directed from negative to positive charge.
- d)Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge.
Correct answer is option ''. Can you explain this answer?
Which of the following statements about dipole moment is not true?
a)
The dimensions of dipole moment is [LTA].
b)
The unit of dipole moment of C m.
c)
Dipole moment is vector quantity and directed from negative to positive charge.
d)
Dipole moment is a scalar quantity and has magnitude charge equal to the potential of separation between charge.
|
|
Geetika Shah answered |
Dipole moment is a vector quantity and has magnitude of 2qa and it is in the direction of the dipole axis from -q to q.
An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.65 x 104 N C-1 in Millikan's oil drop experiment. The density of oil is 1.26 g cm-3. Radius of the oil drop is
(Take, g = 9.8 m s-2, e = 1.6 x 10-19 C)- a)1.04 x 10-6 m
- b)4.8 x 10-5 m
- c)4.8 x 10-18 m
- d)1.13 x 10-18 m
Correct answer is option 'A'. Can you explain this answer?
An oil drop of 10 excess electrons is held stationary under a constant electric field of 3.65 x 104 N C-1 in Millikan's oil drop experiment. The density of oil is 1.26 g cm-3. Radius of the oil drop is
(Take, g = 9.8 m s-2, e = 1.6 x 10-19 C)
(Take, g = 9.8 m s-2, e = 1.6 x 10-19 C)
a)
1.04 x 10-6 m
b)
4.8 x 10-5 m
c)
4.8 x 10-18 m
d)
1.13 x 10-18 m
|
|
Mira Joshi answered |
Here, n = 10, E = 3.65 x 104 N C-1
ρoil = 1.26gcm-3 = 1.26 x 103 kg m-3
As the droplet is stationary weight of droplet = force due to electric field
or (4/3)πr3ρg = neE ∴ r3 = 3neE/4πρg
or r = (1.13 x 10-18)1/3= 1.04 x 10-6 m
ρoil = 1.26gcm-3 = 1.26 x 103 kg m-3
As the droplet is stationary weight of droplet = force due to electric field
or (4/3)πr3ρg = neE ∴ r3 = 3neE/4πρg
or r = (1.13 x 10-18)1/3= 1.04 x 10-6 m
The number of electrons present in -1C of charge is- a)6 x 1018
- b)1.6 x 1019
- c)6 x 1019
- d)1.6 x 1018
Correct answer is option 'A'. Can you explain this answer?
The number of electrons present in -1C of charge is
a)
6 x 1018
b)
1.6 x 1019
c)
6 x 1019
d)
1.6 x 1018
|
|
Riya Banerjee answered |
By quantisation of charge, q = ne
or
or
The dimensional formula of electric flux is- a)[M1L1T-2]
- b)[M1L3T-3A-1]
- c)[M2L2T-2A-2]
- d)[M1L-3T3A1]
Correct answer is option 'B'. Can you explain this answer?
The dimensional formula of electric flux is
a)
[M1L1T-2]
b)
[M1L3T-3A-1]
c)
[M2L2T-2A-2]
d)
[M1L-3T3A1]
|
|
Suresh Iyer answered |
Electric flux, φ= 
The dimension of φ = Dimension of E x dimension of s
[φ] = [M1L1T-2][AT]-1[L2] = [M1L3T-3A-1]
The dimension of φ = Dimension of E x dimension of s
[φ] = [M1L1T-2][AT]-1[L2] = [M1L3T-3A-1]
If the charge on an object is doubled then electric field becomes- a)half
- b)double
- c)unchanged
- d)thrice
Correct answer is option 'B'. Can you explain this answer?
If the charge on an object is doubled then electric field becomes
a)
half
b)
double
c)
unchanged
d)
thrice
|
|
Geetika Shah answered |
As 
If q ' = 2q, then E' =
E' = 2E
So electric field is doubled.
If q ' = 2q, then E' =
E' = 2ESo electric field is doubled.
Two charges q and -3q are fixed on x-axis separated by distance d. Where should a third charge 2q be placed from A such that it will not experience any force?
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Two charges q and -3q are fixed on x-axis separated by distance d. Where should a third charge 2q be placed from A such that it will not experience any force?
a)
b)
c)
d)
|
Lavanya Menon answered |
Let a charge 2q be placed at P, at a distance l from A where charge q is placed, as shown in figure.
The charge 2q will not experience any force, when force of repulsion on it due to q is balanced by force of attraction on it due to -3q at B where AB = d
or
(l + d)2 = 3l2 or 2l2 - 2ld - d2 = 0
If charge 2q is placed between A and B then
Which of the following statement is not a similarity between electrostatic and gravitational forces?- a)Both forces obey inverse square law.
- b)Both forces operate overy very large distances.
- c)Both forces are conservative in nature.
- d)Both forces are attractive in nature always.
Correct answer is option 'D'. Can you explain this answer?
Which of the following statement is not a similarity between electrostatic and gravitational forces?
a)
Both forces obey inverse square law.
b)
Both forces operate overy very large distances.
c)
Both forces are conservative in nature.
d)
Both forces are attractive in nature always.
|
Megha Joshi answered |
Similarities between electrostatic and gravitational forces
Inverse square law
Both electrostatic and gravitational forces obey the inverse square law. This means that the force between two charged or massive objects decreases as the square of the distance between them increases.
Operate over very large distances
Both forces operate over very large distances. The electrostatic force can act between charged particles that are separated by large distances, whereas the gravitational force can act between massive objects that are separated by astronomical distances.
Conservative in nature
Both electrostatic and gravitational forces are conservative in nature. This means that the work done by these forces in moving an object from one point to another is independent of the path taken by the object.
Difference between electrostatic and gravitational forces
Attractive or repulsive
While both electrostatic and gravitational forces are conservative and operate over large distances, the main difference between them is that electrostatic forces can be either attractive or repulsive, depending on the charges of the particles involved. In contrast, gravitational forces are always attractive in nature.
Explanation
The correct answer to the given question is option 'D', as it states that both electrostatic and gravitational forces are always attractive in nature, which is incorrect. Unlike gravitational forces that are always attractive, electrostatic forces can be either attractive or repulsive depending on the charges of the particles involved. This is because like charges repel each other, whereas opposite charges attract each other. Both electrostatic and gravitational forces share the similarities of obeying the inverse square law, operating over large distances, and being conservative in nature.
Inverse square law
Both electrostatic and gravitational forces obey the inverse square law. This means that the force between two charged or massive objects decreases as the square of the distance between them increases.
Operate over very large distances
Both forces operate over very large distances. The electrostatic force can act between charged particles that are separated by large distances, whereas the gravitational force can act between massive objects that are separated by astronomical distances.
Conservative in nature
Both electrostatic and gravitational forces are conservative in nature. This means that the work done by these forces in moving an object from one point to another is independent of the path taken by the object.
Difference between electrostatic and gravitational forces
Attractive or repulsive
While both electrostatic and gravitational forces are conservative and operate over large distances, the main difference between them is that electrostatic forces can be either attractive or repulsive, depending on the charges of the particles involved. In contrast, gravitational forces are always attractive in nature.
Explanation
The correct answer to the given question is option 'D', as it states that both electrostatic and gravitational forces are always attractive in nature, which is incorrect. Unlike gravitational forces that are always attractive, electrostatic forces can be either attractive or repulsive depending on the charges of the particles involved. This is because like charges repel each other, whereas opposite charges attract each other. Both electrostatic and gravitational forces share the similarities of obeying the inverse square law, operating over large distances, and being conservative in nature.
The surface charge density on the outer surface is
- a)-q/4πR21
- b)q/4πR22
- c)q2/4πR21
- d)2q/4πR22
Correct answer is option 'B'. Can you explain this answer?
The surface charge density on the outer surface is
a)
-q/4πR21
b)
q/4πR22
c)
q2/4πR21
d)
2q/4πR22
|
|
Avantika Roy answered |
The given answer, -q/4, is a possible surface charge density on the outer surface of a conductor with charge -q. However, it is not the only possible answer. The surface charge density on the outer surface of a conductor can vary depending on the shape and distribution of the charge.
There are two types of electric charges positive charges and negative charges. The property which differentiates the two types of charges is- a)field of charge
- b)amount of charge
- c)strength of charge
- d)polarity of charge
Correct answer is option 'D'. Can you explain this answer?
There are two types of electric charges positive charges and negative charges. The property which differentiates the two types of charges is
a)
field of charge
b)
amount of charge
c)
strength of charge
d)
polarity of charge
|
|
Swati Chauhan answered |
Explanation:
Electric charges can be classified into two types: positive charges and negative charges. These two types of charges differ in their polarity, which is the property that distinguishes them from each other.
Polarity of Charge:
Polarity refers to the direction of the electric field lines around a charged object. Electric charges are said to have polarity because they create electric fields that have a direction. Positive charges create electric fields that point away from them, while negative charges create electric fields that point towards them.
Thus, the property that differentiates the two types of charges is polarity of charge. Positive charges have a positive polarity, while negative charges have a negative polarity. This property can be observed by the direction of the electric field lines around the charged object.
Conclusion:
Therefore, option D, polarity of charge, is the correct answer.
Electric charges can be classified into two types: positive charges and negative charges. These two types of charges differ in their polarity, which is the property that distinguishes them from each other.
Polarity of Charge:
Polarity refers to the direction of the electric field lines around a charged object. Electric charges are said to have polarity because they create electric fields that have a direction. Positive charges create electric fields that point away from them, while negative charges create electric fields that point towards them.
Thus, the property that differentiates the two types of charges is polarity of charge. Positive charges have a positive polarity, while negative charges have a negative polarity. This property can be observed by the direction of the electric field lines around the charged object.
Conclusion:
Therefore, option D, polarity of charge, is the correct answer.
Two infinite plane parallel sheets, separated by a distance d have equal and opposite uniform charge desities σ. Electric field at a point between the sheets is- a)σ/2ε0
- b)σ/ε0
- c)zero
- d)depends on the location of the point
Correct answer is option 'B'. Can you explain this answer?
Two infinite plane parallel sheets, separated by a distance d have equal and opposite uniform charge desities σ. Electric field at a point between the sheets is
a)
σ/2ε0
b)
σ/ε0
c)
zero
d)
depends on the location of the point
|
|
Dev Patel answered |
Electric field, E = σ/ε0
There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r < R)?- a)E ∝ r-2
- b)E ∝ r-1
- c)E ∝ r
- d)E ∝ r2
Correct answer is option 'C'. Can you explain this answer?
There is a solid sphere of radius R having uniformly distributed charge throughout it. What is the relation between electric field E and distance r from the centre (r < R)?
a)
E ∝ r-2
b)
E ∝ r-1
c)
E ∝ r
d)
E ∝ r2
|
|
Ashish Chaudhary answered |
The electric field inside a uniformly charged solid sphere is given by:
E = 0 for r < />
E = k * (Q / r^2) for r ≥ R
Where:
E is the electric field strength,
k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2),
Q is the total charge of the sphere,
r is the distance from the center of the sphere.
Inside the sphere (r < r),="" the="" electric="" field="" is="" zero="" because="" the="" charges="" cancel="" each="" other="" />
Outside the sphere (r ≥ R), the electric field is inversely proportional to the square of the distance from the center of the sphere. The total charge Q is spread uniformly throughout the sphere, so the electric field follows the same pattern as that of a point charge, where the electric field is given by k * (Q / r^2).
E = 0 for r < />
E = k * (Q / r^2) for r ≥ R
Where:
E is the electric field strength,
k is the electrostatic constant (k = 9 x 10^9 N m^2/C^2),
Q is the total charge of the sphere,
r is the distance from the center of the sphere.
Inside the sphere (r < r),="" the="" electric="" field="" is="" zero="" because="" the="" charges="" cancel="" each="" other="" />
Outside the sphere (r ≥ R), the electric field is inversely proportional to the square of the distance from the center of the sphere. The total charge Q is spread uniformly throughout the sphere, so the electric field follows the same pattern as that of a point charge, where the electric field is given by k * (Q / r^2).
A sphere encloses an electric dipole within it. The total flux across the sphere is - a)zero
- b)half that due to a single charge
- c)double that due to a single charge
- d)dependent on the position of dipole
Correct answer is option 'A'. Can you explain this answer?
A sphere encloses an electric dipole within it. The total flux across the sphere is
a)
zero
b)
half that due to a single charge
c)
double that due to a single charge
d)
dependent on the position of dipole
|
|
Ameya Majumdar answered |
Understanding Electric Flux
Electric flux is a measure of the electric field passing through a given area. According to Gauss's Law, the total electric flux through a closed surface is proportional to the net charge enclosed within that surface.
Electric Dipole Characteristics
- An electric dipole consists of two equal and opposite charges separated by a distance.
- The dipole has no net charge because the positive and negative charges cancel each other out.
Application of Gauss's Law
- Gauss's Law states that the total electric flux (Φ) through a closed surface is given by:
Φ = Q_enclosed / ε₀
- Where Q_enclosed is the total charge within the surface and ε₀ is the permittivity of free space.
Flux Through a Sphere Enclosing a Dipole
- When a sphere encloses an electric dipole, the total enclosed charge (Q_enclosed) is zero.
- Since there is no net charge inside the sphere, the total electric flux through the sphere is also zero.
Conclusion
- The total electric flux across the sphere is zero because the dipole’s positive and negative charges effectively cancel each other out.
- Therefore, the correct answer is option 'A': zero flux across the sphere.
This concept is crucial for understanding how electric fields behave in the presence of dipoles and reinforces the principles of electrostatics as applied in physics.
Electric flux is a measure of the electric field passing through a given area. According to Gauss's Law, the total electric flux through a closed surface is proportional to the net charge enclosed within that surface.
Electric Dipole Characteristics
- An electric dipole consists of two equal and opposite charges separated by a distance.
- The dipole has no net charge because the positive and negative charges cancel each other out.
Application of Gauss's Law
- Gauss's Law states that the total electric flux (Φ) through a closed surface is given by:
Φ = Q_enclosed / ε₀
- Where Q_enclosed is the total charge within the surface and ε₀ is the permittivity of free space.
Flux Through a Sphere Enclosing a Dipole
- When a sphere encloses an electric dipole, the total enclosed charge (Q_enclosed) is zero.
- Since there is no net charge inside the sphere, the total electric flux through the sphere is also zero.
Conclusion
- The total electric flux across the sphere is zero because the dipole’s positive and negative charges effectively cancel each other out.
- Therefore, the correct answer is option 'A': zero flux across the sphere.
This concept is crucial for understanding how electric fields behave in the presence of dipoles and reinforces the principles of electrostatics as applied in physics.
The charge on an electron was calculated by- a)Faraday
- b)J.J. Thomson
- c)Millikan
- d)Einstein
Correct answer is option 'C'. Can you explain this answer?
The charge on an electron was calculated by
a)
Faraday
b)
J.J. Thomson
c)
Millikan
d)
Einstein
|
|
Dev Patel answered |
Charge on an electron was measured by Millikan with the help of the oil drop experiment.
Two large, thin metal places are parallel and close to each other. On their faces, the plates have surface charge densities of opposite signs and of magnitude 16 x 10-22 C m-2. The electric field between the plates is- a)1.8 x 10-10 N C-1
- b)1.9 x 10-10 N C-1
- c)1.6 x 10-10 N C-1
- d)1.5 x 10-10 N C-1
Correct answer is option 'A'. Can you explain this answer?
Two large, thin metal places are parallel and close to each other. On their faces, the plates have surface charge densities of opposite signs and of magnitude 16 x 10-22 C m-2. The electric field between the plates is
a)
1.8 x 10-10 N C-1
b)
1.9 x 10-10 N C-1
c)
1.6 x 10-10 N C-1
d)
1.5 x 10-10 N C-1
|
|
Suresh Iyer answered |
Here 
What will happen when we rub a glass rod with silk cloth?- a)Some of the electrons from the glass rod are transferred to the silk cloth.
- b)The glass rod gets positive charge and silk cloth gets negative charge.
- c)New charge is created in the process of rubbing.
- d)Both (a) and (b) are correct.
Correct answer is option 'D'. Can you explain this answer?
What will happen when we rub a glass rod with silk cloth?
a)
Some of the electrons from the glass rod are transferred to the silk cloth.
b)
The glass rod gets positive charge and silk cloth gets negative charge.
c)
New charge is created in the process of rubbing.
d)
Both (a) and (b) are correct.
|
|
Mira Joshi answered |
When a glass rod is rubbed with silk cloth then some of the electrons from the glass rod get transferred to silk cloth and thus the glass rod gets positive charge and the silk gets negative charge. No new charge is created in the process of rubbing.
A particle of mass 10-3kg and charge 5μC is thrown at a speed of 20m s-1 against a uniform electric field of strength 2 x 105 N C-1. The distance travelled by particle before coming to rest is- a)0.1 m
- b)0.2 m
- c)0.3 m
- d)0.4 m
Correct answer is option 'B'. Can you explain this answer?
A particle of mass 10-3kg and charge 5μC is thrown at a speed of 20m s-1 against a uniform electric field of strength 2 x 105 N C-1. The distance travelled by particle before coming to rest is
a)
0.1 m
b)
0.2 m
c)
0.3 m
d)
0.4 m
|
|
Gaurav Kumar answered |
F = qE = 5 x 10-6 x 2 x 105 = 1 N
Since, the particle is thrown against the field
∴ a = -F/m = -(1/10-3) = 103ms-2
As v2 - u2 = 2as
∴ 02 - (20)2 = 2 x (-103) x s or s = 0.2 m
Since, the particle is thrown against the field
∴ a = -F/m = -(1/10-3) = 103ms-2
As v2 - u2 = 2as
∴ 02 - (20)2 = 2 x (-103) x s or s = 0.2 m
Which of the following figure represents the electric field lines due to a single negative charge?- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
Which of the following figure represents the electric field lines due to a single negative charge?
a)
b)
c)
d)
|
|
Mira Joshi answered |
The field lines of single negative charge are radially inward.
Chapter doubts & questions for Electric Charges and Fields - NCERT Based Tests for NEET 2025 is part of NEET exam preparation. The chapters have been prepared according to the NEET exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for NEET 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Electric Charges and Fields - NCERT Based Tests for NEET in English & Hindi are available as part of NEET exam.
Download more important topics, notes, lectures and mock test series for NEET Exam by signing up for free.
NCERT Based Tests for NEET
684 tests
|
|
© EduRev
|
Education Revolution
|
|
Signup on EduRev and stay on top of your study goals
10M+ students crushing their study goals daily