Force on a Point Charge Due to an Extended Charge

1
Force on a Point Charge Due to an Extended Charge

• Important points to consider
•  
• ? is typically used for charge per length
• Uniform charge means all little pieces of the
  charge are the same.
• Also, uniform charge means
• The limits of integration range from where the
  charge begins to where it ends. There is no
  charge to add-up in the space where there is no
  charge.

2
The Problem A rod, of length 30 cm, is covered
uniformly with 20 ? C of positive charge. The rod
is lying along the x-axis. The rods left end
is 10 cm from the origin. A point charge of 5.0
? C is located at the origin. Find the force felt
by the point charge due to the charged rod.
3
and because charge is uniform,
Let
Now, because the rod lies along the x-axis and
has no height or depth,
because
The expression,
describes the force felt between two point
charges, so we take a little piece of Q, dq, and
treat it as a point charge.
Because we are only asking a small piece of the
question, dq, we get only a small piece of the
answer, dF.
4
Of course, we are usually not satisfied with only
a little of the answer so we integrate (sum up)
all the little pieces of the answer, dF, to get
all of the answer, F. In the spirit of following
rules of algebra, if we sum the left side, we
must sum the right side
Here, r represents the distance from the point
charge to dq and because it lies on the x-axis
Now, as we move along adding up our little pieces
of dF due to the little (dq)s, we will be taking
small steps in the x - direction, dx. That is,
x is what is changing as be go from one dq to
the next dq. Consequently, we must rename dq to
dx.
5
Earlier, we determined
, and now we can rewrite our expression
So,
Pull out the constants
Limits of integration are from the beginning to
the end of the rod
6
Here, the negative sign means repulsion because
the charges have unlike signs. This is NOT
direction. Direction is determined by your
coordinate system.
7
Think about this. We simply substituted values
into an equation and got a negative answer. How
could that represent direction when this equation
is used for all calculations regardless of
direction of the rod and whether the point charge
or the rod is negative? Our answer would have
been the same if the rod had been lying on the
negative axis. Again, your coordinate system
tells the truth about direction. Incidentally,
the electric field, E, points away from the rod.
This is the direction a positive charge would
move if placed near the rod.