Question

The overall goal is to find the electric field at the center of the semicircle below. The charge density is not uniform in this case; it depends on the angle via the function lambda(theta) lambda_0 sin theta. This means that there is more charge density near the center of the semicircle than at the tips. a) If you cut out a small amount of arc length along the circle ds, as shown, then how is ds related to d theta, the amount of angle cut out? b) The total charge on the semicircle is a known number Q. Integrate Q = integral dQ = integral lambda ds over the semicircle in order to obtain a relationship between Q and lambda_0. c) Note that this situation is not symmetric enough to use Gauss's Law, so you must use Coulomb's Law: E = integral kdQ/r^2 r Taking care to analyze the geometry and vector components properly, find E. d) Plug in numbers: if the total charge is Q = 5 mu C and the radius is a = 75cm, what is the field strength?

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Answer 1

lambda(theta) = lambda_0 sin theta a) d theta = ds/a ds = a .d theta b) dQ = lambda ds = lambda_0 sin theta (a d theta) = lambda_0 a sin theta d theta Q = integral d theta = lambda_0 a integral^pi_0 sin theta d theta Q = lambda_0 a [-coa theta]^pi_0 = lambda_0 a [-cos pi + cos 0 degree] Q = lambda_0 a [-(-1)+1] = 2 lambda_0 a Q = 2 lambda_0 a c) d E_Y = k d theta/a^2 sin theta E_y = k/a^2 integral lambda_0 a sin theta d theta sin theta \r\n E_y = k lambda_0/2a integral^pi_0(1-cos 2 theta) d theta E_y = lambda_0K/2a [theta - sin 2 theta/2]^pi_0 E_y = lambda_0k/2a [pi - sin 2pi/2-0+ sin 2(0)/2] E_y = lambda_0 kpi/2a now E_x = integral d E_x = integral^pi_0 K d theta/a^2 cos theta = integral^pi_0 K lambda_0 a sin theta d theta cos theta/a^2 E_x = K lambda_0/2a [- cos 2 theta/2]^pi_0 = K lambda_0/2a 2 [-cos 2pi + cos 0 degree] E_x = K lambda_0/4a 2 = K lambda_0/2a E_x = K lambda_0/2a \r\n E = -E_xi - E_y j i and j are unit vectors along x and y axis