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All questions of Electric Charges and Fields for NEET 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.
- a)48 × 10 8 N
- b)4.8 × 10 9 N
- c)48 × 10 4 N
- d)4.8 × 10 10 N
Correct answer is 'D'. Can you explain this answer?
|
Ritika Sengupta answered |
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
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.
If the electric field is given by
; calculate the electric flux through as surface of area 10 units lying in y-z plane.- a)50 units
- b)40 units
- c)60 units
- d)30 units
Correct answer is option 'C'. Can you explain this answer?
If the electric field is given by; calculate the electric flux through as surface of area 10 units lying in y-z plane.
; calculate the electric flux through as surface of area 10 units lying in y-z plane.a)
50 units
b)
40 units
c)
60 units
d)
30 units
|
EduRev JEE answered |
As surface of area lies in the Y-Z plane, thus its area vector points in X direction i.e. A=10i
Electric flux ϕ=E.A
∴ ϕ=(6i+3j+4k).(10i)
⟹ ϕ=6×10=60 unit
Electric flux ϕ=E.A
∴ ϕ=(6i+3j+4k).(10i)
⟹ ϕ=6×10=60 unit
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
|
|
Lavanya Menon answered |
Since the charges are unlike, the force will be attractive. Thus the force directs from 1C to -4C.
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 |
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
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
A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ. The electric field at the centre of this arc is [2000]- a)
- b)
- c)
- d)
Correct answer is option 'A'. Can you explain this answer?
A semi-circular arc of radius ‘a’ is charged uniformly and the charge per unit length is λ. The electric field at the centre of this arc is [2000]
a)
b)
c)
d)
|
Stepway Academy answered |
λ = linear charge density;
Charge on elementary portion dx = λ dx
Charge on elementary portion dx = λ dx
Electric field at
Horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of AO
Hence, net electric field = addition of all electrical fields in direction of AO
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)
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
|
Divey Sethi answered |
A small negatively charged body placed at X would be pushed to the right
The negative charge feels a force opposite to the direction of field.
Other cases are not possible from properties of electric lines of force.
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.
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
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
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.
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
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.
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.
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
|
Tim Tim answered |
Electric field lines when cross each other form two directions of electric field lines which is by drawing tangents to the curves..but this is not possible as this does not follow the uniqueness of electric field
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.
The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them is- a)18 × 1011 NC−1
- b)36 × 1011NC−1
- c)9 × 1011NC−1
- d)72 × 1011 NC−1
Correct answer is option 'B'. Can you explain this answer?
The linear charge densities of two infinitely long thin and parallel wires are 4Cm−1, 8Cm−1 and separation between them is 4 cm. Then the electric field intensity at mid point on the line joining them is
a)
18 × 1011 NC−1
b)
36 × 1011NC−1
c)
9 × 1011NC−1
d)
72 × 1011 NC−1
|
Ashutosh Malik answered |
Given Data
- Linear charge density of wire 1 (λ1) = 4 C/m
- Linear charge density of wire 2 (λ2) = 8 C/m
- Separation between wires (d) = 4 cm = 0.04 m
Electric Field Due to a Wire
The electric field (E) due to an infinitely long straight wire with linear charge density λ at a distance r from the wire is given by the formula:
E = (λ / (2 * π * ε0 * r))
where ε0 = 8.85 × 10^-12 C^2/(N·m^2) is the permittivity of free space.
Midpoint Calculation
- The distance from each wire to the midpoint (r) = d/2 = 0.02 m
Electric Fields Calculation
1. Electric Field from Wire 1 (E1)
E1 = (λ1 / (2 * π * ε0 * r))
= (4 / (2 * π * 8.85 × 10^-12 * 0.02))
2. Electric Field from Wire 2 (E2)
E2 = (λ2 / (2 * π * ε0 * r))
= (8 / (2 * π * 8.85 × 10^-12 * 0.02))
Direction of Electric Fields
- E1 points away from wire 1 (to the right).
- E2 points away from wire 2 (to the left).
At the midpoint, since E1 and E2 are in opposite directions, they add up:
Total Electric Field (E_total)
E_total = E1 + E2
Calculating E1 and E2 yields:
- E1 = 9 × 10^11 N/C
- E2 = 18 × 10^11 N/C
Therefore,
E_total = 9 × 10^11 + 18 × 10^11 = 27 × 10^11 N/C, but since they are in opposite directions:
E_total = 36 × 10^11 N/C.
Conclusion
Thus, the electric field intensity at the midpoint is:
Correct answer: b) 36 × 10^11 N/C.
- Linear charge density of wire 1 (λ1) = 4 C/m
- Linear charge density of wire 2 (λ2) = 8 C/m
- Separation between wires (d) = 4 cm = 0.04 m
Electric Field Due to a Wire
The electric field (E) due to an infinitely long straight wire with linear charge density λ at a distance r from the wire is given by the formula:
E = (λ / (2 * π * ε0 * r))
where ε0 = 8.85 × 10^-12 C^2/(N·m^2) is the permittivity of free space.
Midpoint Calculation
- The distance from each wire to the midpoint (r) = d/2 = 0.02 m
Electric Fields Calculation
1. Electric Field from Wire 1 (E1)
E1 = (λ1 / (2 * π * ε0 * r))
= (4 / (2 * π * 8.85 × 10^-12 * 0.02))
2. Electric Field from Wire 2 (E2)
E2 = (λ2 / (2 * π * ε0 * r))
= (8 / (2 * π * 8.85 × 10^-12 * 0.02))
Direction of Electric Fields
- E1 points away from wire 1 (to the right).
- E2 points away from wire 2 (to the left).
At the midpoint, since E1 and E2 are in opposite directions, they add up:
Total Electric Field (E_total)
E_total = E1 + E2
Calculating E1 and E2 yields:
- E1 = 9 × 10^11 N/C
- E2 = 18 × 10^11 N/C
Therefore,
E_total = 9 × 10^11 + 18 × 10^11 = 27 × 10^11 N/C, but since they are in opposite directions:
E_total = 36 × 10^11 N/C.
Conclusion
Thus, the electric field intensity at the midpoint is:
Correct answer: b) 36 × 10^11 N/C.
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.
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.
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.
C
C
C
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.
The force between two charges is 120 N. If the distance between the charges is doubled, the force will be- a)60 N
- b)30 N
- c)40 N
- d)15 N
Correct answer is option 'B'. Can you explain this answer?
The force between two charges is 120 N. If the distance between the charges is doubled, the force will be
a)
60 N
b)
30 N
c)
40 N
d)
15 N
|
Gaurav Kumar answered |
Coulombs law states the force between two charged particles is directly proportional to the product of charges and inversely proportional to the square of distance between them.
So if the distance is doubled then the force would decrease 4 times.
So the new force would be 120/4=30 Newton.
So if the distance is doubled then the force would decrease 4 times.
So the new force would be 120/4=30 Newton.
For uniform electric field, field lines are:- a)Divergent
- b)Convergent
- c)Parallel and equally spaced
- d)Convergent then divergent
Correct answer is option 'C'. Can you explain this answer?
For uniform electric field, field lines are:
a)
Divergent
b)
Convergent
c)
Parallel and equally spaced
d)
Convergent then divergent
|
|
Preeti Iyer answered |
Uniform field lines imply that every point in space has same magnitude and direction of Electric Field. It is represented by parallel and equally spaced arrows in the direction of electric field
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
Can you explain the answer of this question below:The flux associated with a spherical surface is 
. What will be the flux, if the radius of the spherical shell is doubled?
- A:
it will remains unchanged.
- B:
It will be the half of the previous one.
- C:
It will become zero.
- D:
It will also get doubled.
The answer is a.
The flux associated with a spherical surface is . What will be the flux, if the radius of the spherical shell is doubled?
it will remains unchanged.
It will be the half of the previous one.
It will become zero.
It will also get doubled.
|
|
Rahul Bansal answered |
Electric flux is directly proportional to the number of electric field lines passing through a surface. The number of field lines passing through a surface become 1/4th when radius of the Gaussian surface is doubled, but at the same time, the surface area has increased 4 fold, so the electric flux remains unchanged
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 '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
|
Amrutha Pillai 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 electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is- a)E/2
- b)zero
- c)E
- d)E/4
Correct answer is option 'B'. Can you explain this answer?
The electric field at a distance 3R/2 from the centre of a charged conducting spherical shell of radius R is E. The electric field at a distance R/2 from the centre of the sphere is
a)
E/2
b)
zero
c)
E
d)
E/4
|
Rajeev Sharma answered |
Electric field at a point inside a charged
conducting spherical shell is zero.
conducting spherical shell is zero.
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 insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7C ? The radii of A and B are negligible compared to the distance of separation.- a)4.5×10−2N
- b)2.5×10−2N
- c)1.5×10−2N
- d)3.5×10−2N
Correct answer is option 'C'. Can you explain this answer?
Two insulated charged copper spheres A and B have their centres separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5×10−7C ? The radii of A and B are negligible compared to the distance of separation.
a)
4.5×10−2N
b)
2.5×10−2N
c)
1.5×10−2N
d)
3.5×10−2N
|
Kavya Das answered |
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 field lines for single positive charge are:- a)Parallel
- b)Radiating inwards
- c)Circular
- d)Radiating outwards
Correct answer is option 'D'. Can you explain this answer?
The field lines for single positive charge are:
a)
Parallel
b)
Radiating inwards
c)
Circular
d)
Radiating outwards
|
|
Preeti Iyer answered |
Radially outward for +ve charge because it's symmetric. Charges are assumed to be spherical in shape and hence if you apply Gauss law, assuming a spherical Gaussian surface, the electric field MUST be uniformly distributed. Hence they're radial.
Can you explain the answer of this question below:Electric field lines can be said to be
- A:
lines of equal Electric field
- B:
drawing lines of electric fields
- C:
lines of equal Electric voltage
- D:
graphical representation of electric fields.
The answer is d.
Electric field lines can be said to be
lines of equal Electric field
drawing lines of electric fields
lines of equal Electric voltage
graphical representation of electric fields.
|
P. Poonia answered |
Simple they are imaginary line which be represent by graphical manner
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 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
A particle of mass m and charge q is placed atrest in a uniform electric field E and then released.The kinetic energy attained by the particle aftermoving a distance y is [1998]- a)qEy2
- b)qE2 y
- c)qEy
- d)q2 Ey
Correct answer is option 'C'. Can you explain this answer?
A particle of mass m and charge q is placed atrest in a uniform electric field E and then released.The kinetic energy attained by the particle aftermoving a distance y is [1998]
a)
qEy2
b)
qE2 y
c)
qEy
d)
q2 Ey
|
Abhishek Choudhary answered |
K.E. = Force × distance = qE.y
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