Two charges are at a distance d apart. If a copper plate of thickness ²⁄_d is kept between them, the effective force will be

F/2
zero
2F
√2F

Solution

The dielectric constant for metal is infinity, the force between the two charges would be reduced to zero.

What is the value of E in the space outside the sheets?

σ/ε_o
σ/2ε_o
E≠0
2σ/ε_o

Solution

Inside a charged conducting surface E= 0, but on or outside the surface E≠0.

∴ Electric intensity is discontinuous across a charged conducting surface.

A charge Q is enclosed by a Gaussian spherical surface of radius R. If the radius is doubled,then the outward electric flux will

increase four times
be reduced to half
remain the same
be doubled

Solution

By Gauss’s theorem, φ = \(\frac{Q_{in}}{\epsilon _{0}}\)

Thus, the net flux depends only on the charge enclosed by the surface. Hence, there will be no effecton the net flux if the radius of the surface is doubled.

If a dipole of dipole moment \(\underset{p}{\rightarrow}\) is placed in a uniform electric field \(\underset{E}{\rightarrow}\), then torque acting on it is given by

\(\underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}\times \underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}+ \underset{E}{\rightarrow}\)
\(\underset{\tau }{\rightarrow}=\underset{p}{\rightarrow}- \underset{E}{\rightarrow}\)

Solution

Given : Dipolemoment of the dipole = \(\underset{p}{\rightarrow}\) and uniformelectric field = \(\underset{E}{\rightarrow}\).Torque (τ) = Either force ×perpendicular distance between the two forces =qaEsinθ or τ=pE sin θ or \(\underset{\tau }{\rightarrow}\) = \(\underset{p}{\rightarrow}\times \underset{E}{\rightarrow}\) (vector form)

The E-r curve for an infinite linear charge distribution will be

Solution

The insulation property of air breaks down when the electric field is 3 × 10⁶ Vm^–1. The maximum charge that can be given to a sphere of diameter 5 m is approximately

2 × 10^–2 C
2 × 10^–3 C
2 × 10^–4C
2 × 10^–5 C

Solution

A charge q is placed at the centre of the open end of a cylindrical vessel. The flux of the electric field through the surface of the vessel is

zero
q/ε_o
q/2ε_o
2q/ε_o

Solution

The flux is zero according to Gauss’ Law because it is a open surface which enclosed a charge q.

Two conducting spheres of radii r₁ and r₂ are charged to the same surface charge density. The ratio of electric fields near their surface is

\(r_{1}^{2}/r_{2}^{2}\)
\(r_{2}^{2}/r_{1}^{2}\)
r₁ / r₂
1 : 1

Solution

In a medium of dielectric constant K, the electric field is \(\overrightarrow{E}\).If ∈₀ is permittivity of the free space, the electric displacement vector is

\(\frac{K \underset{E}{\rightarrow}}{\varepsilon _{0}}\)
\(\frac{\underset{E}{\rightarrow}}{K \varepsilon _{0}}\)
\(\frac{\varepsilon _{0}\underset{E}{\rightarrow}}{K }\)
K ∈₀\(\overrightarrow{E}\)

Solution

The electric field intensity just sufficient to balance the earth’s gravitational attraction on an electron will be: (given mass and charge of an electron respectively are 9.1 × 10 ^-31 kg and 1.6 × 10^-19C.)

–5.6 × 10^–11 N/C
–4.8 × 10^–15 N/C
–1.6 × 10^–19 N/C
–3.2 × 10^–19 N/C

Solution

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