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

Inside the cavity, the net charge is zero, so electric field = 0. Using Gauss's law on the surface of the cavity, we can find the electric field at any point inside the cavity:
E * 4πr^2 = Q/ε0
E = Q/4πε0r^2
Substitute the given values:
E = q/(4πε0r^2)

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

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