A capacitor, C₁= 5.00 μF, is charged up to 8.00 V. It is then connected to a second uncharged capacitor C₂ = 2.50 μF. The voltage across C₂ after the system has come to equilibrium is

Question

Quizplus · Accepted Answer

Before connecting the two capacitors, the total charge stored in the system is $Q = C_1V_1 = (5.00 ext{ μF})(8.00 ext{ V}) = 40.0 ext{ μC}$. When they are connected, the charge will redistribute so that the potential difference across each capacitor is equal. Let $Q_1$ and $Q_2$ be the charge stored in C₁ and C₂ respectively, then we have: $$ Q_1 + Q_2 = Q = 40.0 ext{ μC} $$ $$ V_1 = V_2 $$ Using the capacitance equation, we can write $Q_1$ and $Q_2$ in terms of $V_1$ or $V_2$: $$ Q_1 = C_1V_1 $$ $$ Q_2 = C_2V_2 $$ Substituting $V_1 = V_2$, we have: $$ C_1V_1 + C_2V_1 = Q $$ $$ V_1 = \frac{Q}{C_1+C_2} = \frac{40.0 ext{ μC}}{5.00 ext{ μF} + 2.50 ext{ μF}} = 5.33 ext{ V} $$ Therefore, the voltage across C₂ is 5.33 V.

A Capacitor, C1 = 5

A Capacitor, C₁= 5