Positive charge Q is placed on a conducting spherical shell with inner radius R₁ and outer radius R₂. A particle with charge q is placed at the center of the cavity. The magnitude of the electric field at a point in the cavity, a distance r from the center, is: C) q/4 $\pi$$\varepsilon$₀r² D) (q + Q)/4 $\pi$$\varepsilon$₀r²

Question

Quizplus · Accepted Answer

As the spherical shell is conductive, it creates an electric field inside as well as outside the cavity. However, as there is no charge present inside the cavity, the electric field due to positive charge Q on the outer surface of the shell cancels out the electric field due to negative charge induced on the inner surface of the shell. Hence, the electric field inside the cavity is only due to the charge q present at the center.

By Gauss's Law, we know that the electric field at a distance r from the center of a spherical charge q is given by:
E = q/(4πε0r²)

Here, q = charge of the particle at the center of the cavity and r = distance of the point from the center.

Therefore, E = q/(4πε0r²)

Option C is the only one that matches this expression.

Positive Charge Q Is Placed on a Conducting Spherical Shell π\piπ

Positive Charge Q Is Placed on a Conducting Spherical Shell $\pi$