Irodov 3.117: The Source Adds to the Drop – Amit & Deepali Sinha Foundation

Problem Irodov 3.117

A section $AB$ of a circuit holds, in order from $A$ to $B$ , a capacitor $C_1 = 1.0$ $\mu$ F, a source of emf $\mathcal{E} = 10$ V, and a capacitor $C_2 = 2.0$ $\mu$ F. The source’s negative terminal faces $C_1$ and its positive terminal faces $C_2$ . The potential difference across the section is $\varphi_A - \varphi_B = 5.0$ V. Find the voltage across each capacitor.

The section from A to B: capacitor C_1, the source with its long (+) plate toward C_2, and capacitor C_2. — The section from $A$ to $B$ : capacitor $C_1$ , the source with its long ( $+$ ) plate toward $C_2$ , and capacitor $C_2$ .

Given

$\mathcal{E} = 10$ V
$C_1 = 1.0~\mu\text{F}$
$C_2 = 2.0~\mu\text{F}$
$\varphi_A - \varphi_B = 5.0$ V
middle conductor initially neutral (assumed)

Find

the voltages

V_1

V_2

across

C_1

C_2

Solution

Step 1Both capacitors carry the same charge

The right plate of $C_1$ , the source and the left plate of $C_2$ form one conductor, cut off from the rest of the circuit by the two capacitor gaps. The source moves charge within it but cannot change its total, which we take to be zero, as it was before the section was charged. So if the plate of $C_1$ at $A$ carries $q$ , its other plate carries $-q$ , and the plate of $C_2$ at the source carries $+q$ :

\varphi_A - \varphi_- = \frac{q}{C_1}, \qquad \varphi_+ - \varphi_B = \frac{q}{C_2} \tag{1}

where $\varphi_-$ and $\varphi_+$ are the potentials of the source’s negative and positive terminals.

Step 2Add the potential steps from

A

B

Going from $A$ to $B$ , the potential falls by $q/C_1$ across $C_1$ , rises by $\mathcal{E}$ across the source (from its $-$ terminal to its $+$ terminal), and falls by $q/C_2$ across $C_2$ :

\varphi_A - \varphi_B = (\varphi_A - \varphi_-) - (\varphi_+ - \varphi_-) + (\varphi_+ - \varphi_B) = \frac{q}{C_1} - \mathcal{E} + \frac{q}{C_2}

Solving for the charge:

q = \frac{(\varphi_A - \varphi_B + \mathcal{E})\,C_1C_2}{C_1 + C_2}

(2)

Step 3The voltage across each capacitor

Numerically, the charge (2) is

q = \frac{(5.0 + 10)~\text{V} \times (1.0~\mu\text{F})(2.0~\mu\text{F})}{3.0~\mu\text{F}} = 10~\mu\text{C}

Dividing by each capacitance through (1):

V_1 = \frac{(\varphi_A - \varphi_B + \mathcal{E})\,C_2}{C_1 + C_2} = 10~\text{V}

(3)

V_2 = \frac{(\varphi_A - \varphi_B + \mathcal{E})\,C_1}{C_1 + C_2} = 5.0~\text{V}

(4)

The plate charges: q = 10 on the plate of C_1 at A and on the plate of C_2 at the source, -q on the other two plates. — The plate charges: $q = 10~\mu\text{C}$ on the plate of $C_1$ at $A$ and on the plate of $C_2$ at the source, $-q$ on the other two plates.

$q$ is positive, so the signs are as drawn: the plate of $C_1$ at $A$ and the plate of $C_2$ next to the source’s $+$ terminal are the positive ones.

Check

Walk from $A$ to $B$ with the numbers: down $10$ V across $C_1$ , up $10$ V across the source, down $5.0$ V across $C_2$ , a net fall of $5.0$ V, which is $\varphi_A - \varphi_B$ . The two capacitors carry $C_1V_1 = 10~\mu\text{C}$ and $C_2V_2 = 10~\mu\text{C}$ , equal as Step 1 requires. A special case: with $\varphi_A - \varphi_B = -\mathcal{E}$ the source alone accounts for the whole difference and both capacitors are uncharged, as (2) gives.

Remark

A capacitor in series with a source turns its emf into a fixed offset: the section behaves like one capacitor $C_1C_2/(C_1 + C_2) = 0.67~\mu\text{F}$ charged by the voltage $\varphi_A - \varphi_B + \mathcal{E}$ , as if the source were moved to the end of the section. With the source turned around the same section would see $5.0 - 10 = -5.0$ V, and the capacitors would charge the other way with a third of the charge.