Capacitors of 2 µF and 3 µF are connected in series to 100 V. Find the charge and the voltage on each — Physics problems | PhysLabo
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Two capacitors connected in series

Medium 10 grade Electricity / Electrostatics
Problem statement

Two capacitors of capacitances $C_1 = 2$ µF and $C_2 = 3$ µF are connected in series to a voltage source $U = 100$ V (see the circuit). Find the capacitance of the combination, the charge on each capacitor and the voltage across each of them. The capacitors were uncharged before the connection.

Circuit: two capacitors C₁ and C₂ connected in series to a voltage source U
Given
  • $C_1 = 2$ µF, $C_2 = 3$ µF
  • $U = 100$ V
  • $C = ?$, $q = ?$
  • $U_1 = ?$, $U_2 = ?$
Hints
Hint 1 / 2
For capacitors in series the charge on both is the same and the voltages add up.
Hint 2 / 2
First find the capacitance of the combination $C = \frac{C_1 C_2}{C_1 + C_2}$ and the charge $q = CU$, then the voltages $U_1 = q / C_1$ and $U_2 = q / C_2$.
Solution

For capacitors in series the charge on both is the same (the middle section of the circuit is electrically isolated and stays neutral), and the voltages add up: $U = U_1 + U_2$.

Plate charges and voltages: U = U₁ + U₂
Plate charges and voltages: U = U₁ + U₂

1. The capacitance of the combination:

$$\frac{1}{C} = \frac{1}{C_1} + \frac{1}{C_2} \;\Rightarrow\; C = \frac{C_1 C_2}{C_1 + C_2} = \frac{2 \cdot 3}{2 + 3} = 1.2 \text{ µF}$$

2. The charge (the same on both capacitors):

$$q = CU = 1.2 \cdot 10^{-6} \cdot 100 = 1.2 \cdot 10^{-4} \text{ C} = 120 \text{ µC}$$

3. The voltages:

$$U_1 = \frac{q}{C_1} = \frac{120}{2} = 60 \text{ V}, \qquad U_2 = \frac{q}{C_2} = \frac{120}{3} = 40 \text{ V}$$

Check: $U_1 + U_2 = 100$ V $= U$. The larger voltage is across the capacitor of smaller capacitance.

Answer

$C = 1.2$ µF; $q = 120$ µC; $U_1 = 60$ V, $U_2 = 40$ V

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