A thin glass rod bent in a semi-circular shape symmetric about the x-axis. (See

Question

A thin glass rod bent in a semi-circular shape symmetric about the x-axis. (See Fig. 2.) The quarter circle section above the x axis has a charge of +25.4 micro-Coulomb and the bottom section has a charge of -25.4 micro-Coulomb. The radius of the semi-circular section is is 27.2 cm. You will determine the electric field at the center P by doing the next set of problems. This gives practice in using the steps to solve this type of problem. a) The charge density of the upper quarter circle is ____________. Give your answer in the form "+/-a.bc x 10^(y)" C/m. (Do you agree that the SI unit is C/m? Be sure that you understand that this unit is correct!) b)The charge density of the lower quarter circle is ____________. c)choose a differential segment of length dl on the upper section of the rod. dl is located at an angle of theta degrees counterclockwise from the +y axis. The differential amount of charqe dQ on dl is expressed as dQ = _______ d(theta) . Give your answer in the form "a.bc x 10^(d) unit". Hint. remember that dQ is electric charge so the unit must be the unit for charge. Also, d(theta) must be in radians, not degrees. d)Now perform the integration to complete the problem to find the electric field (magnitude and direction) at P. Give your answer in the form "a.bc x 10^(d)" N/C, "+/-X or Y" axis. Put a comma and one space between the magnitude and the direction. For example: 1.26 x 10^(-4), +X.

Explanation / Answer

a)

a = q/(r) = 25.4e-6/(3.1416*0.272) = 2.9724e-5 +2.97 x 10^-5 C/m

b)

b = q/(r) = (-25.4e-6)/(3.1416*0.272) = -2.9724e-5 -2.97 x 10^-5 C/m

c)

dQ = a dl = a (r d) = r a d

dQ = 0.272 * 2.9724e-5 d = 8.0849 x 10^-6 d 8.08 x 10^-6 d Coulomb

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