Question

Solve part c. What would be the magnitude of the electric field(in N/C) at O if the top half of the semicircle carried a total charge of -7.50 uC and the bottom half, insulated from the top half, carried a total charge of +7.50 uC?

e to the let o dowrward

Answer 1

dl = R d theta dQ = Q/pi r dl = Q/pi R R d theta = Q/pi d theta Taking bottom half circle dE = KdQ/R^2 = K Q/R d theta/R^2 = KQ/pi R^2 d theta Because of Symmetry horizontal compounds will get cancelled implies E = integral^pi_0 d E sin theta = integral^pi_0 kQ/pi R^2 sin theta d theta = KQ/pi R^2 [- Cos theta]^pi_0 = KQ/pi R^2 times 2 = 2KQ/pi R^2 implies E = 2KQ/pi R^2 N/C upward Top half has negative chare therefore field due to this will also be upward and same value implies E' = 2 times e = 2 times 2 KQ/pi R^2 = 4KQ/pi R^2 = 4 times 9 times 10^9 times 7.5 times 10^- 6/pi times (0.17/pi)^2 = 2.93 times 10^2 N/C implies E = 2.93 times 10^3 N/C