Question

a 7.3 - μf capacitor is charged by a 145 - v battery (see figure 1 a) and then is disconnected from the battery. when 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 $Q = C V$. For $C_1=7.3\ \mu F$ and $V = 145\ V$, the initial charge $Q_1 = C_1V=7.3\times10^{- 6}\ F\times145\ V = 1.0585\times10^{-3}\ C$.

Step2: Calculate total capacitance after connection

After connecting $C_1$ and $C_2$, the voltage across both is $V_f = 18\ V$. The total charge $Q_{total}=Q_1$ (charge - conservation). Let the total capacitance be $C_{total}=C_1 + C_2$. Using $Q = C V$, we have $Q_1=(C_1 + C_2)V_f$.

Step3: Solve for $C_2$

Rearranging the equation $Q_1=(C_1 + C_2)V_f$ gives $C_2=\frac{Q_1}{V_f}-C_1$. Substituting $Q_1 = 1.0585\times10^{-3}\ C$, $V_f = 18\ V$ and $C_1=7.3\times10^{-6}\ F$:
\[

$$\begin{align*} C_2&=\frac{1.0585\times 10^{-3}\ C}{18\ V}-7.3\times10^{-6}\ F\\ &=58.8056\times10^{-6}\ F-7.3\times10^{-6}\ F\\ &=51.5056\times10^{-6}\ F\\ &\approx52\ \mu F \end{align*}$$

\]

Answer:

$C_2 = 52\ \mu F$