**Examples
for capacitors**

Example #1

Problem:

Consider two
plates separated by *d*=1.5* cm *, where the
electric field between them is 100 *V/m*, and the charge
on the plates is 30.0 *m**C*. What is the capacitance?

Solution:

The Capacitance is:

2.0E-5
*F*

Example #2

Problem:

Consider a
capacitor made of two 0.05 *m*^{2} plates
separated by 0.5 *mm. *If the capacitance is 3.0 *nf*,
what is the relative permeability, k, of the material between the
plates?

Solution:

3.4

Example #3

Problem: part a.)

What is the capacitance of the following segment of a circuit?

DATA: *C*_{1}*=C*_{2}*=C*_{3}=3.0
*m**f*

Solution:

The capacitance of the two in
parallel is *C*_{23}*=C*_{2}*+C*_{3}*.
*The capacitance of the entire circuit is:

2.0
(*m**F*)

part b.)

If a voltage *V*_{ab}
= 6.0 *V* is applied, what is the charge on each
capacitor?

Solution:

The charge on the first capacitor is the same as the charge on the whole combination, since it is the only thing the left-hand wire is connected to. This charge can be found from the capacitance.

There is a charge *Q*_{1}
on the opposite side of the first capacitor, which must
have come equally from the next capacitors since they are
equal to each other. Therefore,
*Q*_{2}=*Q*_{3}=*(1/2)Q*_{1}.

*Q*_{1}*
= *12 *m**C,
Q*_{2 }= Q_{3
}= 6.0 *m**C*

part c.)

What is the voltage across each capacitor?

Solution:

Use *V*_{i}*=Q*_{i}*/C*_{i}*.
*Check when you are done that the voltages add up to 6.0 *V*.

*V*_{1}*
= *4.0 *V, V*_{2 }=
V_{3 }= 2.0 *V*

Example #4

Problem:

A capacitor has a
charge of 3.0 *nC* when the voltage across the capacitor
is 12 *V*. What is the energy stored in the capacitor?

Solution:

The energy is:

18
E-9 *J*