Physics

Electrostatics

303 Questions

Electrostatics deals with electric charges, fields, and potentials at rest. It is a crucial topic for physics sections in engineering and civil services competitive examinations. Review these questions to build a strong understanding of Coulomb law, electric dipoles, Gauss law, and electric flux.

Electric field and potentialElectric dipole momentGauss Law applicationsCoulomb force calculationsCharge distribution on spheresEquipotential surfaces

Electrostatics Questions

Multiple choice
  1. It is quadrupled.

  2. It is zero.

  3. It is halved.

  4. It is tripled.

  5. It is unchanged.

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The electric potential of a charge is given by the equation V = kq/r. In other words, the distance is inversely proportional to electric potential. If the distance is doubled, then the electric potential must be halved.

Multiple choice
  1. Two connected conductors are at different potentials.

  2. Net charge in any shell is zero.

  3. Charge remains constant in all conductors, except those which are earthed.

  4. Charge on the inner surface of the innermost shell is equal to 0.

  5. Equal and opposite charges appear on opposite faces.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A small sphere will have a lower potential than a large sphere if they both have the same charge on them. Hence, a current will flow when they are connected.

Multiple choice
  1. 1/4 π Є . q/r²

  2. q/4πεor2

  3. 1/4 π Єo Єr . q/r²

  4. P/4πεor3

  5. 4πεor

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Applying Gauss’s Law and Putting values E x 4πr2 = q / ε₀ or E= q / 4πεor2 This expression is same as electric field due to a point charge q placed at distance r,  i.e. in this case if complete charge q is placed at the centre of shell the electric field is same. 

Multiple choice
  1. IV-A, II-B, III-C, I-D

  2. III-A, I-B, II-C, IV-D

  3. II-A, IV-B, I-C, III-D

  4. I-A, III-B, IV-C, II-D

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The correct matching is: Electric force - Newton (force unit), Electric charge - Coulomb (charge unit), Electric Potential - Volt (potential unit), Electric capacity - Farad (capacitance unit). This gives II-A, IV-B, I-C, III-D.

Multiple choice applications of gauss's law coulomb's law physics

A charge is kept at the centre of a shell. Shell has charge Q uniformally distibuted over its surface and radius R. The force on the central charge due to the shell is :

  1. towards left

  2. towards right

  3. upward

  4. zero

Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Electric field at the centre of shell is zero when charge is uniformally distributed over surface of shell . 

Hence the force on the charge at the centre is zero.

Multiple choice applications of gauss's law coulomb's law physics

A hollow conducting sphere of charge does not have electric field at

  1. outer point

  2. interior point

  3. beyond $2m$
  4. beyond $100m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$E=0$, at any point inside the sphere

Multiple choice applications of gauss's law coulomb's law physics

The rupture of air medium occurs at $E=3\times 10^6 \ V/m$. The maximum charge that can be given to a sphere of diameter $5 \ m$ will be (in coulomb):

  1. $2\times 10^{-2}$
  2. $2\times 10^{-3}$
  3. $2\times 10^{-4}$
  4. $2\times 10^{-5}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $E _o= 3 \times 10^6 \ V/m;$  Diameter of sphere $= 5 \ m$

Electric field on the surface of the sphere $= \cfrac {KQ}{R^2}$

$E _o= \cfrac {KQ}{R^2}$

$Q= \cfrac {R^2E^o}{K}= \cfrac {6.25 \times 3 \times 10^6}{9 \times 10^9}$

$= 2 \times 10^{-3}\ C$

Multiple choice applications of gauss's law coulomb's law physics

A positive charge q is placed in a spherical cavity made in a positively charged sphere. The centres of sphere cavity are displaced by a small distance $\overrightarrow l $. Force on charge q is:

  1. in the direction parallel to vector $\overrightarrow l $
  2. in radial direction

  3. in a direction which depends on the magnitude of charge density in sphere

  4. direction can not be determined.

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Inside a uniformly charged sphere, the electric field is uniform and proportional to the displacement vector between the centers of the sphere and the cavity. Therefore, the force on a positive charge placed in the cavity is directed parallel to the displacement vector l.

Multiple choice applications of gauss's law coulomb's law physics

A thinwalled, spherical conducting shell S of radius R is given charge Q. The same amountof charge is also placed at its centre C. Which of the following statements are correct?

  1. On the outer surface of S, the charge density is $\displaystyle \frac {Q} {2 \pi R^2} $
  2. The electric field is zero at all points inside S

  3. At a point just outside S, the electric field is double the field at a point just inside S

  4. At any point inside S, the electric field is inversely proportional to the square of its distance from C

Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

because of charge Q at center there will be induced charge -Q at inner surface of sphere. hence charge density $\dfrac{2Q}{4\pi r^2} = \dfrac{Q}{2\pi r^2}$

because of 2Q charge outside the electric field is double that of inside.
At any point inside S, the electric field is inversely proportional to the square of its distance from C

Multiple choice applications of gauss's law coulomb's law physics

Two sphere's are isolated from each other. They each have an identical net positive charge and have the same radius, however, one sphere is solid and insulating, while the other is a hollow conducting sphere whose charge is uniformly distributed.
For which sphere is the electric field the greatest distance $x$ from the center of the spheres?
Assume $x$ is less than the radius of the spheres.

  1. The conducting hollow sphere has a greater E-field.

  2. The insulating solid sphere has a greater E field.

  3. Both spheres have the same E field.

  4. Neither sphere would cause there to be an Electric field.

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The charge in a conducting hollow sphere resides on the surface only. Hence on applying Gauss Law on a spherical surface enclosed in the sphere, we get electric field to be zero inside it.

For a solid insulating sphere with uniform charge distribution, finite electric field exists in the sphere.
Hence correct answer is option B.

Multiple choice applications of gauss's law coulomb's law physics

As one penetrates through uniformly charged conducting sphere, what happens to the electric field strength:

  1. decreases inversely as the square of the distance

  2. decreases inversely as the distance

  3. becomes zero

  4. increases inversely as the square of distance

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Electric field strength inside the uniform charged sphere is zero.

$\therefore$  As one penetrates through uniformly charged sphere, electric field strength inside the sphere becomes zero.