Question

In the figure, a thin glass rod forms a semicircle of radius r = 4.50 cm. Charge is uniformly distributed along the rod, with +q = 2.50 pC in the upper half and -q = -2.50 pC in the lower half.

(a) What is the magnitude of the electric field at P, the center of the semicircle?( in N/C)

(b) What is its direction? (counterclockwise from the positive x axis)

Answer 1

Explanation:

Q/pi(r)=dq/rd□

dq=Qd□/pi

de=dEcos □

E=integral dE cos□

E=integral kdq/r^2 cos □

E=integral kQcos□d□/pir^2

E=integral kQ/pi(r^2)( cos□ d□)

E=kQ/pir^2 integral cos□d□

E=2kQSin□/2 /pir^2

here □=45

if you superpose both

E=2rt(2)kQsin(□/2) / pir^2