Physics

Electrostatics

299 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

Two metal plates of area 0.01 m2 carries a charge of 100 μC, are separated by a distance of 1 cm. What is the force on metal plates having potential 3 x 106 V?

  1. 3 x 108 V/m

  2. 3 x 104 Vm

  3. 3 x 10-8 m/V

  4. 3 x 106 Vm

  5. 3 x 106 V/m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Distance between the plates - d = 1cm =1×10-2 m We know that, Strength of electric field (E) = Potential(V )/Distance( d) E = 3  x 106/10-2 = 3 x 108 V/m Which is more than dielectric strength of air, so it is not possible.

Multiple choice

Two charges are separated by a distance d. If the distance between them is doubled, what will be the change in electric potential between them?

  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

Which of the following properties is not related to concentric conducting spherical shells?

  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

If V ( = q/4πεor) is the potential at a distance r, due to a point charge q. Calculate the electric field?

  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

Match the following: Quantity Unit A. Electric force I. Volt between two charged bodies B. Electric charge II. Newton C. Electric Potential III. Farad D. Electric capacity IV. Coulomb

  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

When the field behind a screen with an opening is added to the field of a complementary screen, the sum is equal to the field when there is no screen. Which of the following explains the above statement?

  1. Babinet’s principle

  2. Field equivalence principle

  3. Hansen–Woodyard principle

  4. None of these

  5. -
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the field behind a screen with an opening is added to the field of a complementary screen, the sum is equal to the field when there is no screen.

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
D Correct answer