Electric Potential due to a Finite Charged Rodplaza.obu.edu/corneliusk/up2/epfcr.pdf · Ex....
Transcript of Electric Potential due to a Finite Charged Rodplaza.obu.edu/corneliusk/up2/epfcr.pdf · Ex....
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.