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Q.

An ellipsoidal cavity is carved with in a perfect conductor. A positive charge q is placed at the center of the cavity. The points A and B are on the cavity surface as shown in the figure then
a) Electric field near A in the cavity = Electric field near B in the cavity
b) Charge density at A = Charge density at B
c) Potential at A = Potential at B
d) Total electric flux through the surface of the cavity is $q / ε_{0}$ .

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a

a, b, c are correct

b

only a and b are correct

c

only c and d are correct

d

a, b, c, d are correct

answer is D.

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Detailed Solution

Given data:

Charge $= q$

Charge density and electric field

Concepts used: Electric field

Explanation:

The electric field in A is different from that in B. This is because $E = \frac{K}{r^{2}}$

and we know that the conductor is an equipotential surface, so the potentials of A and B are equal. As charge density

$σ \propto \frac{1}{r}$

The charge densities of A and B are different.

Using Gauss' law, flow on the surface of the cavity,

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An ellipsoidal cavity is carved with in a perfect conductor. A positive charge q is placed at the center of the cavity. The points A and B are on the cavity surface as shown in the figure thena) Electric field near A in the cavity = Electric field near B in the cavityb) Charge density at A = Charge density at Bc) Potential at A = Potential at Bd) Total electric flux through the surface of the cavity is q/ε0.