Electrostatics MCQ – Electrostatics Questions and Answers

Answer.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

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.

Answer.3. 2 V/m

Explanation:-

Dielectric constant (ε_r) 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

Answer.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.

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.

Answer.1. 9 × 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 = 1/4πε_o = 9 × 10⁹ Newton

Answer.1. 0.45 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

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

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.

Answer.2. 0.6 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,
q_A = + 4 μC at point A
q_B = – 16 μC at point B
q_C = + 6 μC at point C

Since, net force on charge q_C will zero.

∴ |F_CA| = |F_CB|

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