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

Since there is no charge in the cavity, the electric field inside the cavity is solely due to the positive charge Q on the conducting spherical shell. The electric field inside a spherical shell is zero, so we only need to consider the electric field at the point where the particle is located. By Gauss's law, the electric field at a point a distance r from the center is:
E = Q/(4πε0r^2)
Since the charge q in the cavity does not affect the electric field, we can use the same formula as if the cavity were empty. Thus, the answer is (C).

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

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