# Electrostatics MCQ – Electrostatics Questions and Answers

Ques.1. When the separation between two charges is made four times, the force between them

1. increases four times
2. decreases four times
3. increases sixteen times
4. decreases sixteen times

Explanation:-

The force between two point charges is along the line joining between them is directly proportional to the product of point charges and inversely proportional to the square of the distance between them:

$F \propto \frac{{{Q_1}{Q_2}}}{{{R^2}}}$

When the separation (R) between two charges is made four times, then the force between them is

F ∝ 1/r2

∴ Force is decreased sixteen times

Ques.2. The force per unit length between two stationary parallel wires carrying (steady) currents _____.

A. is inversely proportional to the separation of wires

B. is proportional to the magnitude of each current

C. satisfies Newton’s third law

1. A and B are correct
2. B and C are correct
3. A and C are correct
4. A, B and C are correct

Answer.4. A, B and C are correct

Explanation:-

The force between two current-carrying parallel conductors:

• Two current-carrying conductors attract each other when the current is in the same direction and repel each other when the currents are in the opposite direction.
• Force exerted on one wire due to currents through two wires is inversely proportional to the distance between them. Thus the force becomes half if the distance between the wires is doubled.
• The attractive force between two parallel current-carrying wires is ~ proportional to the product of the two currents.

Force per unit length on conductor

$\frac{F}{l}=~\frac{{{\mu }_{0}}}{2\pi }\frac{{{i}_{1}}{{i}_{2}}}{d}$

It satisfies newton’s third law.

Ques.3. If voltage applied on a capacitor is increased from V to 2V, choose the correct conclusion.

1. Q remains the same, C is doubled
2. Q is doubled, C doubled
3. C remains same, Q doubled
4. Both Q and C remain same

Answer.3. C remains same, Q doubled

Explanation:-

The  relation that relates the capacitance, voltage and charge of a capacitor is

Q = CV

or

Q ∝ V

Here, Q is the charge of the capacitor, C is its capacitance and V is the voltage.

Therefore, we could say the charge in a capacitor is doubled by doubling the voltage.

Hence, option C is found to be the answer.

Ques.4. A dielectric material is placed in vacuum in a uniform electric field of E = 4 V/m. What is the electric field inside the material if the relative permittivity of dielectric material is 2?

1. Zero
2. 4 V/m
3. 2 V/m
4. 8 V/m

Explanation:-

Dielectric constant (εr) is defined as the ratio of the electric permittivity of the material to the electric permittivity of free space.

The presence of the dielectric reduces the effective electric field.

The relative permittivity of the vacuum is 1.

As the relative permittivity increases by a factor of 2, the electric field decreases by a factor of 2.

Therefore, the electric field = 2 V/m

Ques.5. Which of the following law states that “The total flux out of a closed surface is equal to the net charge within the surface”?

1. Laplace’s law
2. Gaussian’s Law
3. Coulomb’s law
4. Gauss’s law

Explanation:-

Gauss law for electric field:

Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.

The net flux of a given electric field through a given surface, divided by the enclosed charge should be equal to a constant.

Differential form: ∇ . E = ρ/ε0

∇ . D = ρ

Gauss law for magnetic field:

Gauss’s law for magnetism is one of the four of Maxwell’s equations. It states that the magnetic field B has divergence equal to zero, in other words, it is a solenoidal vector field.

It is equivalent to the statement that magnetic monopoles do not exist.

Ques.6. The potential inside a charged hollow sphere is __________.

1. Zero
2. Same as that on the surface
3. Less than that on the surface
4. None of these

Answer.2. Same as that on the surface

Explanation:-

• The electric field inside a conducting sphere is zero, so the potential remains constant at the value it reaches the surface.
• When a conductor is at equilibrium, the electric field inside it is constrained to be zero.
• Since the electric field is equal to the rate of change of potential, this implies that the voltage inside a conductor at equilibrium is constrained to be constant at the value it reaches the surface of the conductor.
• A good example is the charged conducting sphere, but the principle applies to all conductors at equilibrium.

Ques.7. What is the force of attraction between two electric charges of opposite polarity having 1 Coulomb each when placed at 1 m?

1. 9 × 109 Newton
2. 5.54 × 1011 Newton
3. 8.857 × 1012 Newton
4. 4π × 109 Newton

Explanation:-

Coulomb’s law: When two charged particles of charges q1 and q2 are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

The force of attraction between two charges is given by

$F = \frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left| {{q_1}{q_2}} \right|}}{{{r^2}}}\:$

Given that, q1 = 1 C, q2 = -1 C, r = 1 m

F = 1/4πεo = 9 × 109 Newton

Ques.8. Two long parallel conductors are placed 10mm apart from each other carrying a current of 150 Amperes. What will be force per meter length of each one?

1. 0.45 N/m
2. 0.1 N/m
3. 4.5 N/m
4. 9 N/m

Explanation:-

Force per unit length between the two conductors is

$\frac{F}{l} = \frac{{{\mu _0}{I_1}{I_2}}}{{2\pi d}}$

Here d = distance = 10 × 10-3 meter

μ0 = 4π × 10-7F/m

And currents I1 = I2 = 150 Ampere

$\therefore \frac{F}{l} = \frac{{4\pi \times {{10}^{ – 7}} \times 150 \times 150}}{{2\pi \times 10 \times {{10}^{ – 3}}}} = 0.45\:N/m$

Ques.9. Which among the following statement is correct regarding an ideal conductor in a static electric field?

1. Static electric field intensity inside conductor is non zero
2. Static field intensity outside conductor is zero
3. Static field intensity at the surface of conductor is directly normal to the surface
4. Static field intensity at the surface of the conductor is directly parallel to the surface

Answer.3. Static field intensity at the surface of conductor is directly normal to the surface

Explanation:-

Under the static condition, the electric field inside the solid perfect conductor is zero in electrostatic equilibrium even if:

• it is isolated
• it is charged
• it is present in the external electrostatic field

Now, it is known that a perfect solid conductor is an equipotential body. The potential inside the conductor is the same as the potential at the surface.

Since the surface of the solid conductor is an equipotential surface hence the electric field will be perpendicular or in other words normal to the perfectly solid conductor surface.

Ques.10. Two charges of + 4 μC and -16 μC are separated from each other by a distance of 0.6 m. At what distance should a third charge of + 6 μC be placed from + 4 μC so that no force exerts on it will be zero?

1. 0.4 m
2. 0.6 m
3. 1.2 m
4. 0.3 m

Explanation:-

Consider new charge + 4μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.

Let,
qA = + 4 μC at point A
qB = – 16 μC at point B
qC = + 6 μC at point C

Since, net force on charge qC will zero.

∴ |FCA| = |FCB|

According to coulomb’s law

$F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}$

$\begin{gathered} \frac{{K{q_A}{q_C}}}{{{d^2}}} = \frac{{K{q_B}{q_C}}}{{{{(0.6 + d)}^2}}} \hfill \\ \hfill \\ \frac{{24}}{{{d^2}}} = \frac{{96}}{{{{(0.6 + d)}^2}}} \hfill \\ \end{gathered}$

4d2 = (0.6 + d)2

4d2 = 0.36 + d2 + 1.2d

3d2 – 1.2d – 0.36 = 0

d1 = – 0.2 m

d2 = + 0.6 m

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