Question

Potential at the edge of disc

Answer 1

(c) 4σkR

Explanation : dV=kdqr

dq=σ(2rθ)dr

dV=14π∈0σ2rdrθr=σ2π∈0θdr

r=2Rcosθ

r=2Rcosθ

dr=−2Rsinθdθ

dr=−σ2π∈02Rθsinθdθ

V=∫0π2dV=σRπ∈0∫π20θsinθdθ

V=σRπ∈0=4kσR.