Ex. Electric Potential due to a Finite Charged Rod Find the electric potential some distance y above a uniformly charged finite rod

x

r

dq

y

P

2 2 2r x y= +

dx

dy

Since this is a uniform charged rod dqdx

λ = dq =

At point P:

V edq

kr

= ∫

2 2V e

L

L

dxk

x y

λ

−

=+∫

Since x2 is an even function, we can change the limits of integratintegral from 0 to L.

2 2

0

2V e

Ldx

kx y

λ=+∫

From integration tables:

( )2 2

2 2ln

dxx x a

x a+ +

+=∫

( ) ( )2 22 lV lnek L L y yλ += + − n

2 2

lnek

y

Q L L yL

+ +=

V Using 2

Q

Lλ = &

Total Linear Charge Density

Total Length of the Rod

22

Q

Ll L

λ =

=

⇒

⇒

dq represents the amount of charge on the rod piece of length dx.

dxλ

ion from –L to L to 2 times the

ln ln lna

a bb

= −

Alternate Integration: We can also determine the electric potential by using the electric field for a finite charged rod.

Recall:

2 2ˆ

2 ek L

y y L

λ=

+yE for a finite charge rod

Using Remember: Our V = 0 point is at Vy

dl∞

= − ⋅∫E r = ∞ (or in this case). y = ∞

2 2

ˆ2

' 'V 'e

yk L

y y Ldyλ

∞ += − ⋅∫ y

2 2

'

' 'V 2 e

ydy

y y Lk Lλ

∞ += − ∫ Note: ˆ ˆ ' cos 'dy dydy θ =⋅ =y y , since 0θ = .

From integration tables: 2 2

2 2

1ln

du a u aa uu au

+ +−

+

=

∫

2 2

ln ln1V 2 (1)e

L L yk LL y

λ= − + + − +

2 2

lnV 2 eL L yk

yλ=

+ +

Note: ln(1) = 0

2 2

lnV e L L yk QL y

+ +=

Using

2

Q

Lλ =

Which is the same result we obtained before.