Question

You have two identical capacitors and an external potential source.
a) Compare the total energy stored in the capacitors when they are connected to the applied potential in series and in parallel.
U(parallel)/U(series) =?
b) Compare the maximum amount of charge stored in each case.
Q(parallel)/Q(series) =?
c) Energy storage in a capacitor can be limited by the maximum electric field between the plates. What is the ratio of the electric field for the series and parallel combinations?
E(parallel)/E(series) =?

Answer 1

Answer:

Explanation:

Given two identical capacitor

i.e they have the same capacitance C

Also let say we have external potential V

a. Now for series connection,

1/Ceq=1/C+1/C

1/Ceq=(1+1)/C

1/Ceq=2/C

Then, Ceq=C/2

Ceq=½C

So the energy stored in a capacitor is given as

Us= ½CeqV²

Us=½×½CV²

Us=¼CV²

Now, for parallel connection

Ceq=C+C

Ceq=2C

Then,

Energy is given as

Up=½CeqV²

Up=½×2CV²

Up=CV²

Comparing this to series

Us=¼CV², since CV²=Up

Then, Us=¼Up

Up=4Us

Up/Us=4

So the energy stored in the parallel capacitor connection is 4 times the energy stored in the series capacitor connection.

b. Charges

For series connection

We know that same charge pass through series connection. Now, for series the potential difference is V/2 since they have the same capacitance, they will share the voltage

And the charge is given as

Q=CeqV

Qs=½CV

Qs=½CV for each capacitor

Then the total charge will be

Qs=½CV×2.

Qs=CV

Now, for parallel connection

The have the same voltage but different charge,

The charge for parallel is given as

Q=CeqV

Qp=2CV

Now comparing

Qp/Qc=2CV / CV

Qp/Qc=2

The maximum charge of the parallel is four times that of the series again

c. Electric field is given as

E=V/d

The potential difference in parallel is V same voltage.

Ep=V/d

Now for series the potential difference is V/2 since they have the same capacitance, they will share the voltage

Es=V/2d.

Then,

Ep/Es=V/d ÷ V/2d

Ep/Es=V/d × 2d/V

Ep/Es=2