# Capacitors in series with dielectrics

1. Jul 26, 2014

### Zondrina

1. The problem statement, all variables and given/known data

A $240 nF$ capacitor and a $4.0 \mu F$ capacitor are connected in series with a $240 V$ potential difference.

The $4.0 \mu F$ capacitor has a parallel plate configuration, with a surface area of $110 cm^2$ and a dielectric material between the plates. The dielectric constant is $\kappa = 8$.

1. Calculate the charge stored in the $4.0 \mu F$ capacitor.

2. Calculate the potential across the $4.0 \mu F$ capacitor and the potential across the $240 nF$ capacitor.

3. Find the spacing between the plates of the $4.0 \mu F$ capacitor.

4. Find the magnitude of the electric field in the $4.0 \mu F$ capacitor.

5. Calculate the energy density in the $4.0 \mu F$ capacitor.

2. Relevant equations

$C_1 = 4.0 \mu F = 4.0 \times 10^{-6} F$
$C_2 = 240 nF = 240 \times 10^{-9} F$

$V = 240 V$
$\kappa = 8$

$q = CV$
$C = \epsilon_0 L$, where $L$ has the dimension of length.

Screenshot of how I figure the circuit looks: http://gyazo.com/6c138b176e22538339aa4ab44b899b50

For series capacitors: $C_{eq} = (\sum_{i=1}^{n} \frac{1}{C_i})^{-1}$

3. The attempt at a solution

1. I'm slightly confused about how to factor in the dielectric material.

A potential is maintained across the circuit and a dielectric is filling the space between the plates of $C_1$. So the effect of the dielectric is to increase the charge $q_1$ on $C_1$, by increasing $C_1$ by factor of $\kappa$?

Using this logic, I could find an equivalent capacitance $C_{12}$ given by:

$C_{12} = ( \frac{1}{\kappa C_1} + \frac{1}{C_2} )^{-1}$

Then using $q_{12} = C_{12} V_{12}$, I could find the charge across $C_{12}$.

Then using the fact that capacitors in series all have the same charge, we could write $q_{12} = q_1 = q_2$.

Does this seem reasonable?

2. Jul 26, 2014

### vela

Staff Emeritus
No, you're given $C_1$. That's the capacitance with the dielectric already factored in.

You need the dielectric constant to answer 3-5, which have to do with the internals of the capacitor.

Last edited: Jul 26, 2014
3. Jul 26, 2014

### Zondrina

Ah I was a bit confused about that. Thank you for you input.

1. An equivalent capacitor is given by:

$C_{12} = ( \frac{1}{C_1} + \frac{1}{C_2} )^{-1} = ( \frac{1}{4.0 \times 10^{-6} F} + \frac{1}{240 \times 10^{-9} F} )^{-1} = 2.2642 \times 10^{-7} = 2.3 \times 10^{-7} F$

The charge on this capacitor is given by:

$q_{12} = C_{12} V_{12} = (2.2642 \times 10^{-7} F) (240 V) = 0.000054 = 5.4 \times 10^{-5} C$

Since the capacitors are in series $q_1 = q_2 = q_{12}$.

Last edited: Jul 26, 2014
4. Jul 26, 2014

### Zondrina

Continuing the work here:

2. To obtain the potential across $C_1$, simply use:

$q_1 = C_1 V_1 \Rightarrow V_1 = \frac{q_1}{C_1} = \frac{5.4 \times 10^{-5} C}{4.0 \times 10^{-6} F} = 13.5 V = 13 V$

To obtain the potential across $C_2$:

$q_2 = C_2 V_2 \Rightarrow V_2 = \frac{q_2}{C_2} = \frac{5.4 \times 10^{-5} C}{240 \times 10^{-9} F} = 225 V = 230 V$

My results seem to be slightly off, not sure why. They should add up to $240 V$.

3. To find the spacing between the plates of $C_1$, simply use the general equation for a capacitor with a dielectric:

$C_1 = \kappa \epsilon_0 L = \kappa \epsilon_0 \frac{A_1}{d_1}$

Hence: $d_1 = \kappa \epsilon_0 \frac{A_1}{C_1} = (8)(8.85 \times 10^{-12} \frac{F}{m}) \frac{0.011 m^2}{4.0 \times 10^{-6} F} = 1.947 \times 10^{-7} m = 1.9 \times 10^{-7} m$

4. Using Gauss' law with a dielectric and noting the electric field is constant between the plates:

$q_1 = \epsilon_0 \oint \kappa \vec{E} \cdot d \vec{A} = \kappa \epsilon_0 \oint \vec{E} \cdot d \vec{A} = \kappa \epsilon_0 E \oint dA$ (Since $\vec{E} \cdot d \vec{A} = E cos(0°) dA$).

The closed surface integral now gives the surface area of the plates, so we write:

$q_1 = \kappa \epsilon_0 E A \Rightarrow E = \frac{q_1}{\kappa \epsilon_0 A} = \frac{5.4 \times 10^{-5} C}{(8) (8.85 \times 10^{-12} \frac{F}{m}) (0.011 m^2)} = 6.934 \times 10^{7} \frac{V}{m} = 6.9 \times 10^{7} \frac{V}{m}$

5. To find the energy density of $C_1$, simply use:

$u = \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} (8.85 \times 10^{-12} \frac{F}{m}) (6.934 \times 10^{7} \frac{V}{m})^2 = 21275.557 \frac{J}{m^3} = 21000 \frac{J}{m^3}$

Do these look good?

Last edited: Jul 26, 2014
5. Jul 27, 2014

### vela

Staff Emeritus
It's just rounding. To the two sig figs you kept, the sum is consistent with 240 V.

You could also use $V=Ed$ since the electric field is uniform. It's a bit simpler, though your calculation is fine too.

You need to account for the dielectric medium here.