Download Physics with Mathematica Fall 2013 Exercise #4 17 Sep 2012

Physics with Mathematica Fall 2013
Exercise #4
17 Sep 2012
Potential and Field from a Uniformly Charged Line Segment
The electrostatic potential from a charge distribution is given, in CGS units, by
V (x) =
�
dq �
|x − x� |
where dq � = ρ(x� )dV for a volume charge density ρ(x), dq � = σ(x� )dA for a surface charge
density σ(x), and dq � = λ(x� )ds for a line charge density λ(x). Given an electrostatic
potential function V (x), the electric field from that charge distribution is E(x) = −∇V (x).
Consider a straight line segment of uniformly distributed charge Q and length L, lying along
the x-axis and centered on the origin. The line charge density is then simply λ = Q/L. Find
the electrostatic potential along the z-axis, that is V (x) = V (0, 0, z). Express your result in
the simplest form that you can, perhaps using the Simplify function in Mathematica. Do
the same for (the z-component of) the electric field along the z-axis.
Test your results by considering the electric field in the limits z � L and z � L, in which
case you ought to be able to use your Physics II knowledge to figure out what you expect.
The best way to find these limits (I think) is to look for the appropriate series expansion in
terms of z. (Use the Documentation Center!) Note that the integral form of Gauss’ Law is
�
E · dA = 4πQencl
Send the grader an email with your notebook as an attachment.