Question

A Semicircular rod is charged uniformly with a total charge Q coulomb. The electric field intensity at the centre of curvature is ………

(A) [(2KQ)/(πR²)] 

(B) [(3KQ)/(πR²)] 

(C) [(KQ)/(πR²)] 

(D) [(4KQ)/(πR²)]

Answer 1

The correct option (A) [(2KQ)/(πR²)]

Explanation:

 Let λ = charge per unit length = (Q/πR)  (1)

charge on slice shown = dq = λR dθ  (2)

Electric field generated by slice = dE = [(K|dq|)/R²]

= [(K|λ|)/R] = dθ from (2)

components of dE : dEx = dE cos θ

dEy = – dE sin θ

Electric field from all slices add up.

∴ Ex = (Kλ/R) ^π∫₀ cos θ dθ = (Kλ/R) [sin θ]^π₀ = (Kλ/R) × 0 = 0

Ey = [(– Kλ)/R] ^π∫₀ sin θ dθ = + (Kλ / R) [cos θ]^π₀ = (Kλ/R)(– 1 – 1)

= [(– 2Kλ)/R]

E = √(Ex² + Ey²) = [(2Kλ)/R]

from (1) E = (2K/R) ∙ (Q/πR) = [(2KQ)/(πR²)]