Question

Charge Q is uniformly distributed along a thin, flexible rod. The rod is then bent into a semicircle of radius R.

Find an expression for the electric potential at the center of the semicircle.

Answer 1

answer)if we make a figure in any way then we can see either x or y component of the field will cancel out  because of the symmetry, so we need to find field along only one direction)

we have the formula for electric fied

E=Kdqsin/R²

we know the formula for linear charge density is given by

=Q/L=dq/ds

dq=ds

ds=Rd

also we have here

2piR=2L

R=L/pi

so now using eqn 1 and replacing the values we get,

E=kds*sin/R²=KRsind/R²=K/Rsind=K/R(-cos)^pi_o

E=K/R(-(-2))

E=2K/R

we have =Q/L

and R=L/pi

E=2KQpi/L²

so answer is

E=2piKQ/L²( if the answer is wrong try -2piKQ/L² because depending on the figure the field can be negative as well)