All Exams >
Class 12 >
Physics Class 12 >
All Questions
All questions of Electric Charges and Fields for Class 12 Exam
Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be
- a)Zero
- b)Along OF
- c)Along OC
- d)None of these
Correct answer is option 'D'. Can you explain this answer?
Six charges are placed at the corner of a regular hexagon as shown. If an electron is placed at its centre O, force on it will be
a)
Zero
b)
Along OF
c)
Along OC
d)
None of these
|
Dr Manju Sen answered |
The charges will be balanced by their counterparts on the opposite side. So, eventually the charges remaining will be 2q and b and 3q on D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
Since the charge distribution is asymmetrical, the net force on charge would be skewed towards D.
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
- a)9.6 x 104 C
- b)1.9 x 105 C
- c)3.8 x 105 C
- d)4.8 x 104 C
Correct answer is option 'B'. Can you explain this answer?
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
a)
9.6 x 104 C
b)
1.9 x 105 C
c)
3.8 x 105 C
d)
4.8 x 104 C
|
Krishna Iyer answered |
- He atom has 2 electrons.
- So, 1 mole of He has 2*N(N is Avogadro's no.) electrons.
- Then total -ve charge in 1 mole He gas is = 2 * N * charge of 1 electron
= 2 * (6.022 * 1023) * (1.6 * 10-19) = 1.92 * 105 C
Can you explain the answer of this question below:A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is- A:0
- B:
- C:
- D:
The answer is b.
A charge q is placed at the centre of the open end of cylindrical vessel whose height is equal to its radius. The electric flux of electric field of charge q through the surface of the vessel is
A:
0
B:
C:
D:
|
Lavanya Menon answered |
Given that, A charge q is placed at the centre of open end Q a cylindrical vessel,we have to find the flux through the surface of the vessel.
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0
so, when charge Q is placed at the centre of open end of a cylindrical vessel then only half of the charge will contribute to the flux, because half will lie inside the surface and half will lie outside the surface.
so, flux through the surface of vessel is q/2ε0
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R
- a)Rλ/ε0
- b)2Rλ/ε0
- c)λ/ε0
- d)none of the above
Correct answer is option 'B'. Can you explain this answer?
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface centred at the origin with r = R
a)
Rλ/ε0
b)
2Rλ/ε0
c)
λ/ε0
d)
none of the above
|
Nandini Patel answered |
The total charge on the body If there are n1 electrons and n2 protons will be |n1-n2|.
Because whichever of the two will be lower in number, will be neautralized by the other and the left electrons/ protons will be the reason for the charge on the body.
; calculate the electric flux through as surface of area 10 units lying in y-z plane.
Can you explain the answer of this question below:Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
- A:
positive
- B:
Zero
- C:
positive or negative depending on which end is grounded
- D:
negative
The answer is b.
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
positive
Zero
positive or negative depending on which end is grounded
negative
|
|
Neha Sharma answered |
The charge on the rod is shared between the rod and the sphere when they are in contact with each other. However, on grounding the charge will flow to the earth and the charge on the sphere becomes zero
The force between two small charged spheres having charges 3 x 10-6C and 4 x 10-6C placed 40 cm apart in air is- a)67.5 x 10-3 N
- b)67.5 x 10-2 N
- c)6.75 x 10-1 N
- d)none
Correct answer is option 'C'. Can you explain this answer?
The force between two small charged spheres having charges 3 x 10-6C and 4 x 10-6C placed 40 cm apart in air is
a)
67.5 x 10-3 N
b)
67.5 x 10-2 N
c)
6.75 x 10-1 N
d)
none
|
Siddharth Tiwari answered |
**Explanation:**
When two charges are placed in close proximity to each other, they exert a force on each other. This force is known as the electrostatic force. The electrostatic force between two charges is governed by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this case, we have a positive charge of 1C and a negative charge of -4C. According to Coulomb's Law, the force between these two charges will be attractive because one charge is positive and the other is negative. The force will act along the line connecting the two charges, from the positive charge to the negative charge.
To determine the direction of the force, we can use a simple mnemonic: "like charges repel, opposite charges attract." Since the charges in this scenario are opposite (positive and negative), they will attract each other. Therefore, the force will act from the positive charge (1C) towards the negative charge (-4C).
Hence, the correct answer is option C: "From 1C to -4C."
When two charges are placed in close proximity to each other, they exert a force on each other. This force is known as the electrostatic force. The electrostatic force between two charges is governed by Coulomb's Law, which states that the force is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
In this case, we have a positive charge of 1C and a negative charge of -4C. According to Coulomb's Law, the force between these two charges will be attractive because one charge is positive and the other is negative. The force will act along the line connecting the two charges, from the positive charge to the negative charge.
To determine the direction of the force, we can use a simple mnemonic: "like charges repel, opposite charges attract." Since the charges in this scenario are opposite (positive and negative), they will attract each other. Therefore, the force will act from the positive charge (1C) towards the negative charge (-4C).
Hence, the correct answer is option C: "From 1C to -4C."
A negative point charge placed at the point A is 
- a) In stable equilibrium along x-axis
- b)In unstable equilibrium along y-axis
- c)in stable equilibrium along y-axis
- d)in unstable equilibrium along x-axis
Correct answer is option 'C,D'. Can you explain this answer?
A negative point charge placed at the point A is
a)
In stable equilibrium along x-axis
b)
In unstable equilibrium along y-axis
c)
in stable equilibrium along y-axis
d)
in unstable equilibrium along x-axis
|
EduRev Support answered |
If the potential energy of the system is minimum, it will be stable equilibrium, i.e , d2U/dx2>0 and when potential energy is maximum then it will be unstable equilibrium, i.e, d2U/dx2<0.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
As along y direction no electric field, potential energy is minimum and it will be stable equilibrium along y-axis.
Along x-axis potential energy is maximum due to all charges situated along x-axis.so it will be unstable equilibrium.
The diagram shows the electric field lines in a region of space containing two small charged spheres, Y and Z then which statement is true?
- a)Magnitude of electric field is same everywhere
- b)Y is negative & Z is positive
- c)Y and Z must have the same sign
- d)A small negatively charged body placed at X would be pushed to the right
Correct answer is option 'D'. Can you explain this answer?
The diagram shows the electric field lines in a region of space containing two small charged spheres, Y and Z then which statement is true?
a)
Magnitude of electric field is same everywhere
b)
Y is negative & Z is positive
c)
Y and Z must have the same sign
d)
A small negatively charged body placed at X would be pushed to the right
|
Sarthak Rajendra Bande answered |
Z is negatively charged particle because field lines terminate to it and as magnitude of number of lines are equal, both ( Y and Z )have opposite charge but equal in magnitude. so X will experience net repulsion i.e. will be pushed to right
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface cantered at the origin with r = R- a)2Rλ/ε0
- b)Rλ/ε0
- c)λ/ε0
- d)none of the above
Correct answer is option 'A'. Can you explain this answer?
A uniform line charge with linear density λ lies along the y-axis. What flux crosses a spherical surface cantered at the origin with r = R
a)
2Rλ/ε0
b)
Rλ/ε0
c)
λ/ε0
d)
none of the above
|
Neha Chauhan answered |
Total charge inside the spherical surface =2Rλ
∫E.dS = qin/ε0
So flux = 2Rλ/ε0
Hence answer is (a)
∫E.dS = qin/ε0
So flux = 2Rλ/ε0
Hence answer is (a)
Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7 C? The radii of A and B are negligible compared to the distance of separation.- a)4.5×10−2N
- b)2.5×10−2 N
- c)1.5×10−2 N
- d)3.5×10−2 N
Correct answer is option 'C'. Can you explain this answer?
Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7 C? The radii of A and B are negligible compared to the distance of separation.
a)
4.5×10−2N
b)
2.5×10−2 N
c)
1.5×10−2 N
d)
3.5×10−2 N
|
Om Desai answered |
Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero- a)10 cm. away from q and between the charge
- b)10 cm. away from 4q and out side the line joining them.
- c)20 cm. away from 4q and between the charge
- d)none
Correct answer is option 'C'. Can you explain this answer?
Two charges 4q and q are placed 30 cm. apart. At what point the value of electric field will be zero
a)
10 cm. away from q and between the charge
b)
10 cm. away from 4q and out side the line joining them.
c)
20 cm. away from 4q and between the charge
d)
none
|
Hansa Sharma answered |
Where E=0 this point is called neutral point.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.
it is the point where electric field of both charge is same
we know tha t E=kQ/r^2
here k=1/4pi€
for 4q charge
let at ''a'' distance we get E=0 which is from q charge
so distance from 4q of 'a' point is 30-a
electric field by 4q charge on a is
E=k4q/(30-a)^2
electric field by q charge on a point
E=kq/a^2
both electric field are equal so put them equal
k4q/(30-a)^2=kq/a^2
solve this we get
4a^2=(30-a)^2
2a=30-a
3a=30
a=10
so at a distance 10cm from charge q we get E=0
distance from 4q charge 30-10=20cm.
A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A particle of mass m and charge Q is placed in an electric field E which varies with time t ass E = E0 sinwt. It will undergo simple harmonic motion of amplitude
a)
b)
c)
d)
|
Preeti Iyer answered |
Due to verifying electric field, it experiences an verifying force :-
F=QE=QE0sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax=QE0
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
mω2A=QE0
⇒A= QE0/mω2
F=QE=QE0sinωt
at maximum amplitude A, it experience a maximum force of:-
Fmax=QE0
also, Restoring force in SHM is given by: - F=mω2x
for amplitude, x=A OR,
mω2A=QE0
⇒A= QE0/mω2
Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1: Q2: Q3, is - a)it is 1:03:05
- b)it is 1:02:03
- c)it is 1:04:09
- d)it is 1:08:18
Correct answer is 'A'. Can you explain this answer?
Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1: Q2: Q3, is
a)
it is 1:03:05
b)
it is 1:02:03
c)
it is 1:04:09
d)
it is 1:08:18
|
Sanaya Kumar answered |
Since the charge goes to outer surface when given inside so the charges on concentri shells are respectively.
Q1, Q1 + Q2, Q1 + Q2 + Q3. Sine their charge densities are equal so σ1 = σ2 = σ3
So Q1 = 3Q2 = 5Q3 so 1 : 3 : 5
Can you explain the answer of this question below:A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to
- A:
x-1
- B:
x-3/2
- C:
x-2
- D:
x5/4
The answer is C.
A small circular ring has a uniform charge distribution. On a far-off axial point distance x from the centre of the ring, the electric field is proportional to
x-1
x-3/2
x-2
x5/4
|
|
Lavanya Menon answered |
Electric field due to a charged ring in given by: -
at point p
∣E∣= KQx /(R2+x2)3/2 Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣E∣= K&x/(x2)3/2=KQ/x2=Eαx−2
so at a large distance, the ring behaves as a point particle.
at point p
∣E∣= KQx /(R2+x2)3/2 Q =λ(2πr)
at a large distance, x≫R, so :- R2+x2≃x2
⇒∣E∣= K&x/(x2)3/2=KQ/x2=Eαx−2
so at a large distance, the ring behaves as a point particle.
Electric field of a point charge extends up to r where 1/r is- a)one
- b)infinity
- c)Zero
- d)two
Correct answer is option 'C'. Can you explain this answer?
Electric field of a point charge extends up to r where 1/r is
a)
one
b)
infinity
c)
Zero
d)
two
|
Sherlin Dsouza answered |
Hii.... so electric extends upto infinity so inverse of infinity is zero... regards SHERLIN
C
C
C
Which one of the following statement regarding electrostatics is wrong ?- a)Charge is quantized
- b)Charge is conserved
- c)There is an electric field near an isolated charge at rest
- d)A stationary charge produces both electric and magnetic fields
Correct answer is 'D'. Can you explain this answer?
Which one of the following statement regarding electrostatics is wrong ?
a)
Charge is quantized
b)
Charge is conserved
c)
There is an electric field near an isolated charge at rest
d)
A stationary charge produces both electric and magnetic fields
|
|
Nandita Ahuja answered |
Explanation:
Electrostatics is the study of electric charges at rest. It deals with the electric forces between charges, the electric field and potential, and the distribution of charges on conductors.
a) Charge is quantized:
The charge on a body or a particle is always a multiple of the elementary charge (1.6 × 10^-19 C). This means that charge is quantized, and we cannot have a fraction of an elementary charge. This is known as the law of conservation of charge.
b) Charge is conserved:
The total charge in a closed system is always conserved. This means that the net charge of a system cannot be created or destroyed; it can only be transferred from one object to another.
c) There is an electric field near an isolated charge at rest:
An isolated charged object creates an electric field around it. This electric field is a vector field that exerts a force on other charged objects in the vicinity of the charged object. The electric field is proportional to the charge and inversely proportional to the distance from the charged object.
d) A stationary charge produces both electric and magnetic fields:
This statement is incorrect. A stationary charge produces only an electric field, not a magnetic field. However, a moving charge produces both electric and magnetic fields.
Conclusion:
Hence, the correct option is (d), which is wrong.
Electrostatics is the study of electric charges at rest. It deals with the electric forces between charges, the electric field and potential, and the distribution of charges on conductors.
a) Charge is quantized:
The charge on a body or a particle is always a multiple of the elementary charge (1.6 × 10^-19 C). This means that charge is quantized, and we cannot have a fraction of an elementary charge. This is known as the law of conservation of charge.
b) Charge is conserved:
The total charge in a closed system is always conserved. This means that the net charge of a system cannot be created or destroyed; it can only be transferred from one object to another.
c) There is an electric field near an isolated charge at rest:
An isolated charged object creates an electric field around it. This electric field is a vector field that exerts a force on other charged objects in the vicinity of the charged object. The electric field is proportional to the charge and inversely proportional to the distance from the charged object.
d) A stationary charge produces both electric and magnetic fields:
This statement is incorrect. A stationary charge produces only an electric field, not a magnetic field. However, a moving charge produces both electric and magnetic fields.
Conclusion:
Hence, the correct option is (d), which is wrong.
The Gaussian surface for a point charge will be- a)Cube
- b)Cylinder
- c)Sphere
- d)Cuboid
Correct answer is option 'C'. Can you explain this answer?
The Gaussian surface for a point charge will be
a)
Cube
b)
Cylinder
c)
Sphere
d)
Cuboid
|
Rahul Bansal answered |
The electric field of a sphere of uniform charge density and total charge charge Q can be obtained by applying Gauss' law. Considering a Gaussian surface in the form of a sphere at radius r > R, the electric field has the same magnitude at every point of the surface and is directed outward.
Quarks are entities that have- a)Fractional charges, combination of these are present in protons and neutrons
- b)integer charges, combination of these are present in protons, electrons and neutrons
- c)fractional charges, combination of these are present in protons and electrons
- d)no charge, combination of these are present in protons and electrons
Correct answer is 'A'. Can you explain this answer?
Quarks are entities that have
a)
Fractional charges, combination of these are present in protons and neutrons
b)
integer charges, combination of these are present in protons, electrons and neutrons
c)
fractional charges, combination of these are present in protons and electrons
d)
no charge, combination of these are present in protons and electrons
|
Prabhakar Kumar Tiwari answered |
(a)quarks combine to form hadrons most of which are protons and neutrons.quarks are the only known particles whose charges are not integral multiples of their charges u can refer Google for detailed information
If electric field lines cross each other that would mean- a)the electric field can be in either of the two directions.
- b)the direction of the field changes at that point
- c)two directions for the electric field at one point which is not possible.
- d)the resultant electric field at a point is zero
Correct answer is option 'C'. Can you explain this answer?
If electric field lines cross each other that would mean
a)
the electric field can be in either of the two directions.
b)
the direction of the field changes at that point
c)
two directions for the electric field at one point which is not possible.
d)
the resultant electric field at a point is zero
|
Sanika Shaikh answered |
Electric field lines can neither be intersect nor be meet it means it has two direction at single point.
In space of horizontal EF(E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position. 
- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
In space of horizontal EF(E = (mg)/q) exist as shown in figure and a mass m attached at the end of a light rod. If mass m is released from the position shown in figure find the angular velocity of the rod when it passes through the bottom most position.
a)
b)
c)
d)
|
Priyanka Sharma answered |
According to the work energy theorem we have
We+Wg=(1/2)mv2
We have work done by electrostatic force as
We=qElsinθ
and work done by the gravitational force as
Wg=mg(l−lcosθ)
Thus we get
qElsinθ+mg(l−lcosθ)=(1/2)mv2
Thus we get
mgsinθ+mgl−mglcosθ=(1/2)mv2
as θ=45o, we get
mgl=(1/2)mv2
also as v=ωl
we get
ω=√2g/l
We+Wg=(1/2)mv2
We have work done by electrostatic force as
We=qElsinθ
and work done by the gravitational force as
Wg=mg(l−lcosθ)
Thus we get
qElsinθ+mg(l−lcosθ)=(1/2)mv2
Thus we get
mgsinθ+mgl−mglcosθ=(1/2)mv2
as θ=45o, we get
mgl=(1/2)mv2
also as v=ωl
we get
ω=√2g/l
In SI units, a unit of charge is called a- a)ampere
- b)milli coulomb
- c)milli ampere
- d)Coulomb
Correct answer is option 'D'. Can you explain this answer?
In SI units, a unit of charge is called a
a)
ampere
b)
milli coulomb
c)
milli ampere
d)
Coulomb
|
Ayush Joshi answered |
The coulomb (symbolized C) is the standard unit of electric charge in the International System of Units (SI). It is a dimensionless quantity, sharing this aspect with the mole. A quantity of 1 C is equal to approximately 6.24 x 1018, or 6.24 quintillion.
In terms of SI base units, the coulomb is the equivalent of one ampere-second. Conversely, an electric current of A represents 1 C of unit electric charge carriers flowing past a specific point in 1 s. The unit electric charge is the amount of charge contained in a single electron.
The field lines for single negative charge are:- a)Radiating outwards
- b)Radiated inwards
- c)Parallel
- d)Spheres concentric with charge
Correct answer is option 'B'. Can you explain this answer?
The field lines for single negative charge are:
a)
Radiating outwards
b)
Radiated inwards
c)
Parallel
d)
Spheres concentric with charge
|
Anoushka Chopra answered |
<b>Field lines for a single negative charge</b>
<b>Explanation:</b>
When considering a single negative charge, the field lines have certain characteristics that can be observed and understood. The correct answer to the question is option 'B', which states that the field lines for a single negative charge radiate inwards. This can be explained in detail as follows:
<b>1. Definition of field lines</b>
- Field lines are a visual representation of the electric field surrounding a charged particle. They indicate the direction and strength of the electric field at different points in space.
<b>2. Characteristics of field lines</b>
- Field lines always start from positive charges and end on negative charges.
- The density of field lines represents the strength of the electric field. Closer field lines indicate a stronger field, while sparser field lines indicate a weaker field.
- Field lines are continuous curves that never intersect each other.
<b>3. Field lines for a single negative charge</b>
- When considering a single negative charge, the field lines start from the charge and extend outward.
- However, since the question asks for the field lines of a single negative charge, the correct answer is option 'B' - the field lines radiate inward towards the charge.
- This is because the field lines originate from imaginary positive charges that would be attracted towards the negative charge.
<b>4. Visualization of field lines</b>
- To visualize the field lines for a single negative charge, one can imagine placing positive test charges at various points around the negative charge.
- The positive test charges will experience a force of attraction towards the negative charge and will follow the path of the electric field lines.
- By observing the path followed by the positive test charges, one can trace the field lines that radiate inward towards the negative charge.
In conclusion, the field lines for a single negative charge radiate inward towards the charge. This indicates the direction and strength of the electric field surrounding the negative charge.
<b>Explanation:</b>
When considering a single negative charge, the field lines have certain characteristics that can be observed and understood. The correct answer to the question is option 'B', which states that the field lines for a single negative charge radiate inwards. This can be explained in detail as follows:
<b>1. Definition of field lines</b>
- Field lines are a visual representation of the electric field surrounding a charged particle. They indicate the direction and strength of the electric field at different points in space.
<b>2. Characteristics of field lines</b>
- Field lines always start from positive charges and end on negative charges.
- The density of field lines represents the strength of the electric field. Closer field lines indicate a stronger field, while sparser field lines indicate a weaker field.
- Field lines are continuous curves that never intersect each other.
<b>3. Field lines for a single negative charge</b>
- When considering a single negative charge, the field lines start from the charge and extend outward.
- However, since the question asks for the field lines of a single negative charge, the correct answer is option 'B' - the field lines radiate inward towards the charge.
- This is because the field lines originate from imaginary positive charges that would be attracted towards the negative charge.
<b>4. Visualization of field lines</b>
- To visualize the field lines for a single negative charge, one can imagine placing positive test charges at various points around the negative charge.
- The positive test charges will experience a force of attraction towards the negative charge and will follow the path of the electric field lines.
- By observing the path followed by the positive test charges, one can trace the field lines that radiate inward towards the negative charge.
In conclusion, the field lines for a single negative charge radiate inward towards the charge. This indicates the direction and strength of the electric field surrounding the negative charge.
Can you explain the answer of this question below:If mica and woolen cloth are rubbed together, then mica gets- A:positively charged
- B:negatively charged
- C:remains neutral
- D:dual charged
The answer is a.
If mica and woolen cloth are rubbed together, then mica gets
A:
positively charged
B:
negatively charged
C:
remains neutral
D:
dual charged
|
Mira Sharma answered |
When Mika (quartz) and wooden cloth are rubbed together then Mika gets Positively charged. It's because of friction. Due to friction one gets Positively charged i.e Mika and other gets Negatively charged i.e Wooden cloth.
A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then- a)(Q/q) = (4/1)
- b)(Q/q) = (2/1)
- c)(Q/q) = (3/1)
- d)(Q/q) = (5/1)
Correct answer is option 'B'. Can you explain this answer?
A certain charge Q is divided at first into two parts, (q) and (Q-q). Later on the charges are placed at a certain distance. If the force of interaction between the two charges is maximum then
a)
(Q/q) = (4/1)
b)
(Q/q) = (2/1)
c)
(Q/q) = (3/1)
d)
(Q/q) = (5/1)
|
|
Om Desai answered |
The force between q & Q-q is given by :-
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1
F = kq(Q-q)/r²
As it is given that force between these two is Maximum....so it will be max. when dF/dq is zero
i.e, dF/dq = {k(Q-2q)}/r² = 0
so, Q = 2q
Q/q = 2/1
Can you explain the answer of this question below:At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?
- A:
at the surface of spherical shell
- B:
outside the spherical shell
- C:
inside the spherical shell
- D:
at the centre of spherical shell
The answer is a.
At what point is the electric field intensity due to a uniformly charged spherical shell is maximum?
at the surface of spherical shell
outside the spherical shell
inside the spherical shell
at the centre of spherical shell
|
Bittu Raj Bittu answered |
This is because of when a charge given to a hollow sphere that is spherical shell then all charges will reside on its surface and as we know that electric field is directly proportional to given charge . so according to this concept we can say about this point.
Can you explain the answer of this question below:The electric field intensity due to a sphere (solid or hollow) at an external point varies as-
- A:
1/r
- B:
does not depend upon r
- C:
1/r3
- D:
1/r2
The answer is d.
The electric field intensity due to a sphere (solid or hollow) at an external point varies as-
1/r
does not depend upon r
1/r3
1/r2
|
Agnal Jose answered |
It followw inverse square law as distance of observational point from source charge increases the electric field intensity decreases hence option d is correct
When a negatively charged conductor is connected to earth,
- a)No charge flow occurs.
- b)Protons flow from the conductor to the earth.
- c)Electrons flow from the earth to the conductor.
- d)Electrons flow from the conductor to the earth.
Correct answer is option 'D'. Can you explain this answer?
When a negatively charged conductor is connected to earth,
a)
No charge flow occurs.
b)
Protons flow from the conductor to the earth.
c)
Electrons flow from the earth to the conductor.
d)
Electrons flow from the conductor to the earth.
|
|
Riya Banerjee answered |
Explanation:
When a negatively charged conductor is connected to the earth, electrons will flow from the conductor to the earth. This is because electrons have a negative charge and they will be repelled from the negatively charged conductor and attracted to the positively charged earth. As electrons flow from the conductor to the earth, the negative charge on the conductor will gradually decrease until it becomes neutral.
- Option A is incorrect because charge flow does occur when a negatively charged conductor is connected to the earth.
- Option B is incorrect because protons have a positive charge and they are not free to move in a conductor.
- Option C is incorrect because electrons flow from the earth to the conductor, not the other way around.
If an excess charge is placed on an isolated conductor, then, that amount of charge- a)gets neutralized.
- b)resides on the surface of conductor.
- c)either resides on the surface of conductor or gets neutralized.
- d)move inside the conductor
Correct answer is option 'B'. Can you explain this answer?
If an excess charge is placed on an isolated conductor, then, that amount of charge
a)
gets neutralized.
b)
resides on the surface of conductor.
c)
either resides on the surface of conductor or gets neutralized.
d)
move inside the conductor
|
|
Anaya Patel answered |
An isolated conductor is any metal object not connected to or not in contact with any other conductor . one of its common properties is that there is no charge of any nature within the surface of the conductor. All charge, if any, always resides on the outer surface only.thus , if we place any amount of charge, it would tend to reside on its surface.
Ratio of the permittivity of medium to the permittivity of free space is known as
- a)Dielectric ratio
- b)Dielectric permittivity
- c)Dielectric constant
- d)Dielectric medium
Correct answer is option 'C'. Can you explain this answer?
Ratio of the permittivity of medium to the permittivity of free space is known as
a)
Dielectric ratio
b)
Dielectric permittivity
c)
Dielectric constant
d)
Dielectric medium
|
|
Hansa Sharma answered |
The ratio of a medium's permittivity to the permittivity of free space is known as the dielectric constant or relative permittivity. The equation for relative permittivity is κ = ϵ/ϵ0, where ϵ is the permittivity of the medium and ϵ0 is the permittivity of free space. Both permittivity and the permittivity of free space have the same unit, farads per meter (F/m), so the dielectric constant is dimensionless.
The Gaussian surface for a line charge will be- a)Sphere
- b)Cylinder
- c)Cube
- d)Cuboid
Correct answer is option 'B'. Can you explain this answer?
The Gaussian surface for a line charge will be
a)
Sphere
b)
Cylinder
c)
Cube
d)
Cuboid
|
|
Rajeev Saxena answered |
Electric Field of Line Charge. The electric field of an infinite line charge with a uniform linear charge density can be obtained by a using Gauss' law. Considering a Gaussian surface in the form of a cylinder at radius r, the electric field has the same magnitude at every point of the cylinder and is directed outward.
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.- a)positive
- b)Zero
- c)positive or negative depending on which end is grounded
- d)negative
Correct answer is option 'B'. Can you explain this answer?
Charging a metal sphere by contact using a positively charged rod, followed by grounding can result in ________ charge in a metal sphere.
a)
positive
b)
Zero
c)
positive or negative depending on which end is grounded
d)
negative
|
|
Preeti Iyer answered |
The charge on the rod is shared between the rod and the sphere when they are in contact with each other. However, on grounding the charge will flow to the earth and the charge on the sphere becomes zero
A charged particle of charge q and mass m is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time 't' seconds is- a)
- b)
- c)
- d)
Correct answer is option 'B'. Can you explain this answer?
A charged particle of charge q and mass m is released from rest in an uniform electric field E. Neglecting the effect of gravity, the kinetic energy of the charged particle after time 't' seconds is
a)
b)
c)
d)
|
Vivek Rana answered |
Force on particle=F=qE
Hence, acceleration of particle=a=F/m=qE/m
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m
Kinetic energy=K=(1/2)mv2=(1/2)m(qEt/m)2
⟹K=E2q2t2/2m
Hence, acceleration of particle=a=F/m=qE/m
Initial speed=u=0
Hence, final velocity=v=u+at=qEt/m
Kinetic energy=K=(1/2)mv2=(1/2)m(qEt/m)2
⟹K=E2q2t2/2m
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
The total negative charge in 1 mol of helium (atomic number 2, atomic mass 4) is:
a)
b)
c)
d)
|
Isha Rane answered |
He atom has 2 electrons. so 1 mole of He has 2*N(N is Avohadro's no.) electrons. then total -ve charge in 1 mole He gas is 2*N*charge of 1 electron. calculating we will get about 1.9*10^5 C.
A point charge of 2.0 μCμC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?- a)1.7×105Nm2/C
- b)2.1×105Nm2/C
- c)2.2×105Nm2/C
- d)1.9×105Nm2/C
Correct answer is option 'C'. Can you explain this answer?
A point charge of 2.0 μCμC is at the centre of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
a)
1.7×105Nm2/C
b)
2.1×105Nm2/C
c)
2.2×105Nm2/C
d)
1.9×105Nm2/C
|
Princy Goyal answered |
This question is based on Gauss law. Net electric flux will be 1/Enot times the total charge enclosed
The quantization of charge has little practical consequence and can be ignored at the- a)microscopic level
- b)macroscopic level
- c)all levels
- d)information given is insufficient to decide
Correct answer is 'B'. Can you explain this answer?
The quantization of charge has little practical consequence and can be ignored at the
a)
microscopic level
b)
macroscopic level
c)
all levels
d)
information given is insufficient to decide
|
|
Roshni Desai answered |
Gain or loss of charge in terms of few electrons each carrying very small charge is unobservable at macroscopic level.
A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -- a)For all values of H
- b)Only if H >
- c)Only if H <
- d)Only if H =
Correct answer is option 'B'. Can you explain this answer?
A charged particle having some mass is resting in equilibrium at a height H above the centre of a uniformly charged non-conducting horizontal ring of radius R. The force of gravity acts downwards. The equilibrium of the particle will be stable -
a)
For all values of H
b)
Only if H >
c)
Only if H <
d)
Only if H =
|
|
Hansa Sharma answered |
electric field due to non-conducting ring along its axis is given by,
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
E=kqz/(Z2+r2)3/2
where z is the separation between point of observation to the centre of the ring, q is the charge on the ring and r is the radius of the ring.
We know, electric field will be maximum only when z = R/√2, meaning at this point the value of force must be maximum. and if we increase the value of z from R/√2 it will decrease and if we decrease the value of z from R/√2 , it will increase.
condition of stable equilibrium,
A small vertical displacement upwards should cause the resultant force on the particle to be downwards, to return it or a small vertical displacement downward should cause the resultant force on the particle to be downward.
hence, at z > R/√2 or, h > R/√2 system will be in stable equilibrium.
When a negatively charged conductor is connected to earth,- a)No charge flow occurs.
- b)Protons flow from the conductor to the earth
- c)Electrons flow from the earth to the conductor
- d)Electrons flow from the conductor to the earth
Correct answer is option 'D'. Can you explain this answer?
When a negatively charged conductor is connected to earth,
a)
No charge flow occurs.
b)
Protons flow from the conductor to the earth
c)
Electrons flow from the earth to the conductor
d)
Electrons flow from the conductor to the earth
|
|
Om Desai answered |
After earthing a positively charged conductor electrons flow from earth to conductor and if a negatively charged conductor is earthed then electrons flows from conductor to earth.
Two identical point charges are placed at a separation of l.P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve ?- a)
- b)
- c)
- d)
Correct answer is option 'D'. Can you explain this answer?
Two identical point charges are placed at a separation of l.P is a point on the line joining the charges, at a distance x from any one charge. The field at P is E. E is plotted against x for values of x from close to zero to slightly less than l. Which of the following best represents the resulting curve ?
a)
b)
c)
d)
|
|
Vivek Rana answered |
At point P,
E=E1−E2 rightward
⟹E=(kq/x2)−(kq/(l−x)2) where k=1/4πϵo
⟹E=kq (l(l−2x)/x2(l−x)2)
From graph we can see that it can't be straight line and E=0 at x=l/2
At x=0 ,,E→∞ and at x=l, E→−∞
Hence, from the shown graphs the correct answer is (D).
Earthing or grounding means- a)placing the apparatus with an insulating stand, on the ground
- b)connecting to a green colored wire
- c)sharing the charges with the earth
- d)using green color to paint the body
Correct answer is option 'C'. Can you explain this answer?
Earthing or grounding means
a)
placing the apparatus with an insulating stand, on the ground
b)
connecting to a green colored wire
c)
sharing the charges with the earth
d)
using green color to paint the body
|
|
Preeti Iyer answered |
Earthing and Grounding are actually different terms for expressing the same concept. Ground or earth in a mains electrical wiring system is a conductor that provides a low impedance path to the earth to prevent hazardous voltages from appearing on equipment. In electrical engineering, ground or earth is the reference point in an electrical circuit from which voltages are measured, a common return path for electric current, or a direct physical connection to the earth.
Two positive charges - a)repel each other
- b)attract each other at times and repel at other times
- c)attract each other
- d)always attract each other
Correct answer is option 'A'. Can you explain this answer?
Two positive charges
a)
repel each other
b)
attract each other at times and repel at other times
c)
attract each other
d)
always attract each other
|
Ramesh Chand answered |
In contrast to the attractive force between 2 objects with opposite charges , two objects that are of like charge will repel eachother. That is , a positively charged object will exert a repulsive force upon a second positively charged object. This repulsive force will push the two object apart.
Chapter doubts & questions for Electric Charges and Fields - Physics Class 12 2025 is part of Class 12 exam preparation. The chapters have been prepared according to the Class 12 exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Class 12 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Electric Charges and Fields - Physics Class 12 in English & Hindi are available as part of Class 12 exam.
Download more important topics, notes, lectures and mock test series for Class 12 Exam by signing up for free.
Physics Class 12
105 videos|345 docs|99 tests
|
Signup to see your scores go up within 7 days!
Study with 1000+ FREE Docs, Videos & Tests
10M+ students study on EduRev
