Potential due to a uniformly charged finite line on its axis
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axis between y = -a and y = +a. A point p lies at x along x-axis. Find the electric potential at point P.
Linear charge density:
λ=
Q
2a
Small element of charge:
dQ=λdy
dV in terms of linear charge density:
dV =
dQ
4πϵ0r
=
λdy
4πϵ0
√
x2+y2
Integrate from -a to a by using the integral in integration table, specifically ∫
dx
√
a2+x2
=ln(x+
√
a2+x2
),
V =
λ
4πϵ0
a
∫
−a
dy
√
x2+y2
=
λ
4πϵ0
ln(
√
a2+x2
+a
√
a2+x2
–a
)
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