## PHYS208 Fundamentals of Physics II

### Quiz 3 -- Charge Distributions

Charge + *Q* is distributed uniformly along a line segment of length *a*.

- Derive an expression for the electric field at a point
*P* a distance *r*
from the end of the segment.
- Demonstrate that for points far away from the segment, this expression for the
electric field behaves like that of a point charge.

### Take-Home Component

A similar segment of charge - *Q* is joined to the segment of + *Q* as shown below.

- Derive an expression for the electric field at a point
*P* a distance *r*
from the center of the charge distribution.
- Demonstrate that for points far away from the charge distribution, this expression for the
electric field behaves like that of a point
**dipole**.

### Solution for Takehome Component:

The solution for the left-hand side of the charge distribution, the positively-charged
part was solved as the in-class part of this quiz.
(See solution.)

pointing to the right.

The negatively-charged part will be considered separately and then added to the result above
invoking the superposition principle.

#### Step 1: Understand the geometry

Same as positively-charged part.

#### Step 2: Span the charge distribution

Same as positively-charged part, except that the integration variable *x* is defined
to be positive to the right as shown.
The entire charge distribution may again be spanned by varying *x* from 0 to *a*.
Alternatively, the variable *u* with an origin at the point of evaluation *P*
may be selected, with limits of integration from *r* to *r* - *a*.

Assuming a uniform charge distribution about the ring,
the linear charge density *lambda* will again be defined as *Q* / a.
The negative sign of the charge will be handled here via the direction of the electric field
contribution and not by defining a negative charge density.

#### Step 3: Evaluate the contribution from the infinitesimal charge element

The infinitesimal contribution to the electric field is

#### Step 4: Exploit symmetry as appropriate

Nothing to do...

#### Step 5: Set up the integral

Substitution of variable from *x* to *u* = *r* - *x*
has been made to facilitate solution of the integral.

#### Step 6: Solve the integral

#### Step 7: The result!

Again, for *r* >> *a* this expression reduces to the expected result
for a point charge *Q*. Keep in mind that the charge is now negative so this electric
field points to the left, rather than to the right as that from the positively-charge piece.

#### Step 8: Superposition

**E**_{-} is larger than **E**_{+}, as expected since the
negative charge is closer to the point if evaluation. The magnitude of the
net electric field will thus be **E**_{-} - **E**_{+}, to the left.

#### Step 9: Check Far-Field Limit

As expressed above, the limit may be checked without the binomial theorem.
As *a* becomes small compared to *r*, the denominator becomes *r*^{3}
and the result behaves as a point dipole of dipole moment *p* = *Qa*,
the charge times the "center of mass" separation.
Note that the binomial theorem could be applied with success to the second line above,
as demonstrated here.

"http://www.physics.udel.edu/~watson/phys208/quiz3soln.html"

Last updated Mar. 11, 1998.

Copyright George Watson, Univ. of Delaware, 1998.