Irodov 3.102: Filling One Capacitor Weakens Its Field

Problem Irodov 3.102

Two identical parallel-plate air capacitors, each of capacitance $C$ , are joined in series and connected to a battery of emf $\mathcal{E}$ . One of the two is then completely filled with a uniform dielectric of permittivity $\varepsilon$ . By what factor does the electric field inside the filled capacitor decrease, and how much charge passes through the battery?

The battery of emf E and the two equal air capacitors C in series. — The battery of emf $\mathcal{E}$ and the two equal air capacitors $C$ in series.

Given

two capacitors $C$ in series
emf $\mathcal{E}$
one filled with $\varepsilon$
cell connected (assumed)

Find

the factor

E_0/E

in the filled capacitor · the charge

\Delta q

through the battery

Solution

Step 1Before the filling

In series both capacitors carry the same charge, and their voltages add up to the emf. Two equal capacitances share it equally, so each has the voltage $\mathcal{E}/2$ and the charge

q_0 = C\cdot\frac{\mathcal{E}}{2} = \frac{C\mathcal{E}}{2} \tag{1}

With $d$ the plate separation, the field between the plates of either one is uniform:

E_0 = \frac{\mathcal{E}}{2d} \tag{2}

Step 2After the filling

Filling a capacitor completely with the dielectric multiplies its capacitance by $\varepsilon$ , so the pair is now $\varepsilon C$ in series with $C$ . The battery stays connected throughout (assumed: the question about the charge through it implies it), so the two voltages still add up to $\mathcal{E}$ , and the common charge is the emf times the series capacitance:

q = \mathcal{E}\,\frac{\varepsilon C\cdot C}{\varepsilon C + C} = \frac{\varepsilon C\mathcal{E}}{\varepsilon + 1} \tag{3}

After the filling: the filled capacitor ε C and the air capacitor C carry the same charge q, with the plate facing the cell's positive terminal positive. — After the filling: the filled capacitor $\varepsilon C$ and the air capacitor $C$ carry the same charge $q$ , with the plate facing the cell’s positive terminal positive.

The filled capacitor’s voltage is its charge over its capacitance, and its field is that voltage over the same separation $d$ :

\begin{gathered} V_\varepsilon = \frac{q}{\varepsilon C} = \frac{\mathcal{E}}{\varepsilon + 1} \\ E = \frac{V_\varepsilon}{d} = \frac{\mathcal{E}}{(\varepsilon + 1)d} \end{gathered}

Step 3How many times the field dropped

Dividing (2) by this field:

\dfrac{E_0}{E} = \dfrac{\varepsilon + 1}{2}

(4)

Step 4The charge through the battery

The charge that passes through the battery is the change of the capacitors’ common charge, (3) minus (1):

\Delta q = \frac{\varepsilon C\mathcal{E}}{\varepsilon + 1} - \frac{C\mathcal{E}}{2} = C\mathcal{E}\,\frac{2\varepsilon - (\varepsilon + 1)}{2(\varepsilon + 1)}

so that

\Delta q = \dfrac{C\mathcal{E}}{2}\cdot\dfrac{\varepsilon - 1}{\varepsilon + 1}

(5)

It is positive for $\varepsilon > 1$ : the battery drives this charge from its negative terminal to its positive one, adding to the charge on the plates.

Check

The voltages still add up to the emf: the filled capacitor holds $\mathcal{E}/(\varepsilon + 1)$ and the air one $q/C = \varepsilon\mathcal{E}/(\varepsilon + 1)$ , and the two sum to $\mathcal{E}$ . With $\varepsilon = 1$ nothing changes: the ratio (4) is 1 and $\Delta q = 0$ . With $\varepsilon \to \infty$ the filled capacitor acts as a wire, so the field in it is suppressed entirely and the air capacitor takes the whole emf with the charge $C\mathcal{E}$ ; then $\Delta q \to C\mathcal{E}/2$ , which matches (5).

Remark

The field in the air capacitor rises in the same proportion as the charge, by the factor $2\varepsilon/(\varepsilon + 1)$ , up to twice its old value. Putting a dielectric into one element of a series chain makes the remaining elements more likely to break down, not less: the stress is moved, not removed.