Question

P1. Consider a uniformly-charged rod of length 5.0 cm. 2.5 cm past the end of this rod, the electric field is measured to be 45 kN/C, pointing away from the rod. Calculate the charge on the rod.

Answer 1

lambda = Q/L Small electric field due to small charge dE = k dq/x^2 = k lambda dx/x^2 E = integral^d + L_d k lambda dx/x^2 = -k lambda Vertical-bar 1/x Vertical-bar ^d + L_d = -k lambda Vertical-bar 1/d + L - 1/d Vertical-bar = K lambda [1/d - 1/d + L] = k Q/L (d + L - d/d(d + L)] = k Q/d(d + L) \r\n E = 45 kN/c = 45 times 10^3 N/c we know, electric field due to rod E = kQ/d(d + L) here, d = 2.5 times 10^-2 m L = 5 times 10^-2, m, so, 45 times 10^3 = 8.99 times 10^9 times Q/0.025 (0.05 + 0.025) Q = 45 times 10^3 times 0.025 times 0.075/8.99 times 10^9 - 9.375 times 10^-9 C Q = 9.38 n C