Question

2) In the below semicircular distribution of charges, the upper left quarter of the cirele is positive λ-+20 nC/m, while the upper right quarter of the circle is negative λ- -15 nC/m. The circle radius R - 25 cm. calc. the following at point P, the center of circle (a) the x component of the electric field at point P due to the upper left quadrant charges. Just get the magnitude of this component (its positive value). (b) the x component of the electric field at point P due to the upper right quadrant charges. Just get the magnitude of this component (its positive value)

Answer 1

lambda_1 = 20ne/m = 20 times 10^-9 e/m R = 25cm = 0.25m lambda_2 = -15 ne/m = -15 times 10^-9 e/m a) therefore E_x = E cos theta Let us assume a small portion on the are thatis dl dl = R d theta therefore charge on that portion, dv = lambda_1 dl = lambda_1 R d theta therefore electric field at p due to this small portion dE^+ = 1/4pi epsilon_0 lambda_1 R d theta/R^2 or, dE^+ = 1/4pi epsilon_0 lambda_1 d theta/R therefore dE^+_x = dE cos theta = 1/4pi epsilon_0 lambda_1 cos theta d theta/R therefore E^+_x = integral dE_x = 1/4pi epsilon_0 integral^pi/2_0 lambda_1 cos theta d theta/R = 1/4pi epsilon_0 lambda_1 /R integral^pi/2_0 cos theta d theta = 1/4pi epsilon_0 lambda_1 /R [+ sin theta]^pi/2_0 = 1/4pi epsilon_0 lambda_1 /F times 1 = 1/4pi epsilon_0 lambda_1 /R therefore E^+_x = 9 times 10^9 times 20 times 10^-9 /0.25 [As 1/4pi epsilon_0 = 9 times 10^9] implies E^+_x = 720 N/c b) similarly electric