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
Answers
Answered by
0
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
Similar questions