# Electric Potential due to a Finite Charged Rod ```Ex. Electric Potential due to a Finite Charged Rod
Find the electric potential some distance y above a uniformly charged finite rod
r =x +y
2
P
λ=
dy
2
Q
2L
l = 2L
2
⇒ Total Linear Charge Density
⇒ Total Length of the Rod
r
y
dq
x
Since this is a uniform charged rod &AElig;
λ=
dq represents the amount of charge
on the rod piece of length dx.
dx
dq
dx
dq = λ dx
&Icirc;
At point P:
V = ke
dq
∫r
L
V = ke
λ dx
∫
x +y
2
−L
2
Since x2 is an even function, we can change the limits of integration from –L to L to 2 times the
integral from 0 to L.
L
V = 2λ k e
dx
∫
x +y
2
0
2
From integration tables:
dx
∫
&Icirc;
x +a
2
(
V = 2λ ke  ln L +

V=
k eQ
L
2
(
= ln x + x 2 + a 2
L +y
2
2
 L + L2 + y 2 

y


ln 
)
) − ln ( y )
Using λ =
a
&amp; ln   = ln a − ln b
2L
b
Q
Alternate Integration:
We can also determine the electric potential by using the electric field for a finite charged rod.
Recall:
E=
2ke λ L
y y +L
2
y
2
yˆ
for a finite charge rod
K
∫
∞
Using V = − E ⋅ dl
y
V=−
∫
Remember: Our V = 0 point is at r = ∞ (or y = ∞ in this case).
2ke λ L
∞ y'
K
y' + L
2
y
V = −2keλ L
∫
∞ y'
2
yˆ ⋅ dy '
K
dy '
Note: yˆ ⋅ dy = yˆ dy ' cosθ = dy ' , since θ = 0 .
y' + L
2
2
From integration tables:
∫
&Icirc;
 a + u2 + a2 
= − ln 

a 
u
u u2 + a2

du
1
 1  L + L2 + y 2 

V = −2keλ L  − ln 
 + ln(1) 

y
 L 


 L + L2 + y 2 
V = 2keλ ln 



y


Note: ln(1) = 0
 L + L2 + y 2
V=
ln 

L
y

Using λ =
keQ




Which is the same result we obtained before.
Q
2L
```