A uniformly charged rod of total charge +Q is bent into a half circle of radius

Question

A uniformly charged rod of total charge +Q is bent into a half circle of radius a as shown in the picture.

a) fint the linear density in terms of Q and a.
b) write the equation for the field, dE, at the origin due to the charge, dQ, which is on a rod segment of lenght ds=a d(theta).( express the variable of integration dQ in terms of d(theta):dQ = (?) d(theta))
c)Now express E at the origin as an integral of dE. Include the limits of integration.
d) Perform the integration to find the electric field, E, at the origin due to the charge on the rod. Your answer should be in terms of Q and a, and some constants.

Make sure to show your work, explain your reasoning and label the figure. Q 3. A uniformly charged rod of total charge +Q is bent into a half circle of radius a as shown in the figure da a) Find the linear charge density, , in terms of Q and a b) write the equation for the field, dE, at the origin due to the charge, dQ, which is on a rod segmentof length ds = a d8. (Express the variable of integration dQ in terms of dedas(?)de c) Now express E at the origin as an integral of dE. Include the limits of integration. d) Perform the integration to find the eiectric field, E, at the origin due to the charge on the rod. Your answer should be in terms of Q and a, and some constants.

Explanation / Answer

3. given a uniformly charged rod, total charge Q, is bent into a semicircle of radius a

a. charge density , lambda = Q/length of rod
   length of rod = pi*a
   so lambda = Q/pi*a

b. electric field at origin due to the given segment of charge dQ
   dE = k*dQ/a^2 [ where k is coloumbs constant]
   now, dQ = lambda*ds
   but ds = a*d(theta)
   so, dE = k*lambda*a*d(theta)/a^2 = k*lambda*d(theta)/a

c. the electric field dE is along the line joining the segmet to the center
   but from symmetry, when all of the rod is considered, the component of electric field perpendicular to the axis shown cancel out
   so only the component along the axis adds up
   so E = integral of [k*lambda*cos(theta)*d(theta)/a] from theta = -90 to theta = 90

d. E = k*lambda*[sin(90) + sin(90)]/a = 2k*lambda/a
   lambda = Q/pi*a
   so E = 2k*Q/pi*a^2

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