25.35

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Serway 25.35
A rod of length L lies along the x axis with its left end at the origin. It has a nonuniform
charge density λ = αx, where α is a positive constant.
(a) What are the units of α?
(b) Calculate the electric potential at A.
Here's a new one for you. Usually we only concern ourselves with uniform charge
distributions. But, in this problem, the charge density changes as you move along the x
axis. This makes finding the electric potential a little more difficult, but not much.
(a) The units of α can be found by algebraically satisfying the units of λ, given the
relationship λ = αx. You know the units of x, they're meters. You also know that a linear
charge density, like λ, must have units of coulombs per meter. So, plugging in units
rather than numbers...
  x
C
 m
m

C
m2
(b) Let's start with our general formula for finding the electric potential at some point
with respect to a charge distribution.
(1) V  k e 
dq
r
Our first task is to parameterize dq in terms of something that has spatial meaning.
Well, each little infinitesimal charge, dq, resides within an infinitesimal space along the x
axis, which we'll call dx. The amount of charge in each little chunk of x axis, dx, just
depends on the charge density at that location, so:
dq  dx .
But, since this problem is a little more advanced, λ changes based upon how far down
the x axis you are:
dq  dx  xdx .
Okay, so we now have dq in a form we can use for integration. Now, we need to rewrite
the 'r' in Eqn. (1) so we can use it in the integration, too. 'r' just stands for the distance
from point A to each little chunk of charge. Since A is located a distance d away from
the origin on the negative x axis, and since the location of each little chunk dq is
variable, we can write r as:
rdx
Plugging back into Eqn. (1), and establishing limits of integration to account for all the
charge, we obtain:
dq
V  ke 
 ke
r
V  ke 
x f L

xi 0
dq
 k e
r
xdx
dx
x f L
xdx
dx
xi 0

To solve the integral, set u  d  x . Then, xdx  u  d du .
V  k e 
u  d du  k
u
 du  d  u du   k u
1
e
e
uf
ui
 d ln u  u f
Plugging back in for u:

V  k e d  x  0  d ln d  x  0
L
L


 L 
V  k e L  d ln d  L   ln d   k e  L  d ln 1  
 d 


 L 
V  k e  L  d ln 1  
 d 

u
i

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