University:
Santa Fe College
Course:
ETS 2527C | Electromechanical Components and Mechanisms
Academic year:

2024
Views:

378

Pages:

4
Author:

Jabin F.

# Cavity in a conductor:- All charges in different points of cavity remains independent to each other. E_cavity ≠ 0 E_sphere = 0 Ques.2 P-11 q_enc = +2 E⃗ = (1 / 4πε₀r²) r̂ q_a = 0 q_b = +q q_c = -q q_d = +2q (i) (r < a), q_enc = 0, E⃗ = 0 (ii) (a < r < b), q_enc = 0, E⃗ = 0 (iii) b < r < c, q_enc = φ, E⃗ = (φ / 4πε₀r²) r̂ (iv) c < r < d, q_enc = φ + φ = 0, E⃗ = 0 (v) r > d, q_enc = φ + 2φ, E⃗ = (2φ / 4πε₀r²) r̂ Sol: σ_r=b = σ_r=d φ / (4πb²) = 2φ / (4πd²) φ = √(2)b V_centre|_{r=0} = (1 / 4πε₀) [ (2φ / d) - (φ / c) + (φ / b) ] Ques. 12 P-13 a < r < b V_p = (1 / 4πε₀) [ (2φ / b) - (φ / r) + (φ / b) ] V_p = (φ / 4πε₀b) Ques. 12 P-13 a < r < b V_p = (1 / 4πε₀) [ (2φ / b) - (φ / r) + (φ / b) ] V_p = (φ / 4πε₀b) (option-c) Ques. 19 P-14 V|_{r=b} = V₀ = (1 / 4πε₀) [ (2φ / b) - (φ / b) + (φ / b) ] V₀ = (φ / 4πε₀b) outer V|_{r=b} = V₀ = (1 / 4πε₀) [ (2φ / b) - (φ / b) + (φ / b) ] V₀ = (φ / 4πε₀b) q' = 4πε₀V₀b # r > b q_enc = 2φ - 2φ + q' = q' E = (q' / 4πε₀r²) = (4πε₀V₀b) / (4πε₀r²) = (V₀b) / (r²) r̂ V = (1 / 4πε₀) [ (2φ / r) - (φ / r) + (φ / b) ] = (V₀b) / r Now: Ques → V₀ = 0, q' = 0 V|_{r=0} = (1 / 4πε₀) [ (2q / R) - (2q / a) + (q' / b) ] = (1 / 4πε₀) [ (2q / R) - (2q / a) ] Let us consider that outer surface of shell is connected to a battery of potential V₀, then charge on the outer surface of outer shell is q'.