Question

In the shown figure
ring is bent in the 2 o circular arc with radius R and charge Q what is the electric field E and potential due to rod​

Answer 1

Answer:

Now,

λ=

L

Q

λ=

πR

Q

dq=λdx

=

πR

Q

dx

Potential at P

∫dV

1

=∫

R

Kdq

=∫

Rπ

KQ

R

dx

=∫

0

πR

πR

Q

×

4πε

0

R

dx

V

p

=

4πε

0

R

Q

V

p

=

4πε

0

R

λ×πR

V

p

=

4ε

0

λ