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Electric Charges and FieldsNEET Physics · Class 12 · NCERT Chapter 1

31 NEET previous-year questions on Electric Charges and Fields, each with the correct answer and a step-by-step solution. Filter by topic and expand any question to see how to solve it.

PYQ frequency · topic × year

17
18
19
20
21
22
23
24
Coulomb's law
2
1
1
Field of point charge
1
1
1
Superposition
1
1
1
Dipole
1
2
1
1
1
Electric flux
1
1
1
1
Gauss's law
1
1
1
1
Line / plane
1
1
1
Sphere / shell
2
1
1

Darker = more questions in our PYQ bank for that topic and year.

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All (31)
Coulomb's law (4)
Field of point charge (3)
Superposition (3)
Dipole (6)
Electric flux (4)
Gauss's law (4)
Line / plane (3)
Sphere / shell (4)

A

Repulsive,

B

Attractive,

C

Zero

D

Repulsive,

Solution

Opposite charges attract. Magnitude is (the negative sign just means attraction).

A

B

C

D

Solution

Coulomb's law in field form: for a point charge.

A

B

C

D

Solution

Axial: . Equatorial: .

A

B

C

D

Solution

Total flux = . By symmetry, each of 6 faces gets .

A

B

C

Zero

D

Solution

Gauss's law: enclosed charge inside shell = 0, so flux = 0 and E = 0 everywhere inside.

A

B

C

D

Solution

From Gauss's law (cylindrical surface): .

A

B

C

D

Solution

, where is the angle between E and the area normal.

A

Zero

B

C

D

Solution

The two attractive forces on -q are equal in magnitude and opposite in direction. They cancel: net force = 0.

A

B

C

D

Solution

Infinite plane: . Note that this does NOT depend on distance.

A

B

C

D

Solution

, magnitude .

A

Halved

B

Doubled

C

Quadrupled

D

Same

Solution

Flux depends only on the enclosed charge, not on the surface size: , unchanged.

A

Zero

B

C

D

Solution

Each pulls toward itself with force . The angle between these two forces is , so by parallelogram rule, magnitude .

A

B

C

D

Solution

. Minimum at .

A

Maximum

B

Minimum

C

Zero

D

Cannot say

Solution

By Gauss's law, . No charge enclosed .

A

B

C

D

Zero

Solution

Inside solid sphere, enclosed Q . By Gauss, .

A

Same

B

C

D

Solution

. Charge , distance , so , i.e., 1/8 of original.

A

0.45 N

B

4.5 N

C

22.5 N

D

45 N

Solution

.

A

B

C

D

Solution

Equatorial field: . Half of axial.

A

Zero

B

C

D

Solution

By symmetry, the four field vectors at the centre cancel pairwise. Net field = 0.

A

Zero

B

C

D

Solution

Flux through the hemisphere = flux through its flat circular base (since the closed surface has zero net flux from the uniform field): .

A

B

C

Zero

D

Solution

Outside the shell behaves like a point charge at the centre: .

A

Only spherical surfaces

B

Only conductors

C

Any closed surface

D

Only static charges

Solution

Gauss's law is exact for ANY closed surface. Spherical or cubical or irregular, the relation holds.

A

B

C

D

Solution

In a medium, .

A

Zero

B

C

D

Solution

, with : . Maximum.

A

Zero

B

C

D

Solution

At small , the y-components add (approximately linearly), x-components cancel. .

A

B

C

D

Solution

Infinite line: . Doubling r halves E.

A

B

C

D

Zero

Solution

Inside the outer shell, the outer charge contributes 0 (shell theorem). Only matters. .

A

Zero

B

C

D

Solution

.

A

V

B

V/m

C

V·m

D

C/m²

Solution

has units of (V/m) m² = V·m. Equivalent to N·m²/C.

A

Zero

B

C

D

Solution

Symmetric: forces from the two charges are equal and opposite, so net = 0. (q₀ at the midpoint of two equal charges.)

A

C/(N·m)

B

C²/(N·m²)

C

N·m/C

D

N/(C·m)

Solution

From : has units , equivalent to F/m.

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