Question

a 7.9 μf capacitor is charged by a 145 v battery (see figure 1 a) and then is disconnected from the battery. this capacitor (c1) is then connected (see figure 1 b) to a second (initially uncharged) capacitor, c2. the final voltage on each capacitor is 18 v. part a what is the value of c2? hint: charge is conserved. express your answer using two significant figures and include the appropriate units.

Explanation:

Step1: Calculate initial charge on $C_1$

The formula for charge on a capacitor is $Q = C V$. Given $C_1=73\ \mu F$ and $V = 145\ V$, so $Q_1 = C_1V=73\times10^{- 6}\times145\ C$.

Step2: Calculate total capacitance after connection

After connection, the two - capacitors are in parallel and the potential difference across them is $V_f = 18\ V$. The total charge $Q = Q_1$ (charge conservation). Let the total capacitance be $C_{total}=C_1 + C_2$. And $Q = C_{total}V_f=(C_1 + C_2)V_f$.

Step3: Solve for $C_2$

We know $Q_1=(C_1 + C_2)V_f$. So $C_1V=(C_1 + C_2)V_f$. Rearranging for $C_2$ gives $C_2=\frac{C_1(V - V_f)}{V_f}$. Substituting $C_1 = 73\times10^{-6}\ F$, $V = 145\ V$ and $V_f = 18\ V$:
\[

$$\begin{align*} C_2&=\frac{73\times10^{-6}(145 - 18)}{18}\\ &=\frac{73\times10^{-6}\times127}{18}\\ &\approx5.1\times10^{-4}\ F = 510\ \mu F \end{align*}$$

\]

Answer:

$510\ \mu F$