A charge q is uniformly distributed over a disc of radius R. A circular hole of radius R/2 is punched out, taking one radius of the original disc as the diameter of the hole. The potential at the centre of the original disc due to the remaining part is nV₀, where V₀ = [q/(2πε₀R)](1 − 1/π). Find n.
Physics · Electrostatic
JEE Mains Advanced

Question

A charge q is uniformly distributed over a disc of radius R. A circular hole of radius R/2 is punched out, taking one radius of the original disc as the diameter of the hole. The potential at the centre of the original disc due to the remaining part is nV₀, where V₀ = [q/(2πε₀R)](1 − 1/π). Find n.

Physics · Electrostatic

G1 Physics · Accepted Answer

Step 1: Potential at the centre of a uniformly charged disc is proportional to ∫dq/r. Removing the offset circular hole is handled by superposition: potential of full disc minus potential of the removed disc. Step 2: Carrying out the radial/angular integration gives exactly Vremaining=[q/(2πε₀R)](1−1/π)=V₀. Step 3: Therefore n=1.