Question

a 73 - 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. figure 1 of 1 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 $C_1=73\ \mu F$ and $V = 145\ V$, the initial charge $Q_1 = C_1V=73\times10^{- 6}\times145\ C$.

Step2: Calculate total capacitance after connection

When the two capacitors are connected, the total charge $Q$ is conserved and the voltage across both capacitors is the same ($V_f = 18\ V$). Let the capacitance of the second - capacitor be $C_2$. The total capacitance $C_{total}=C_1 + C_2=(73\times10^{-6}+C_2)$. The total charge $Q$ (which is equal to the initial charge on $C_1$) is also given by $Q = C_{total}V_f=(73\times10^{-6}+C_2)\times18$. Since $Q = C_1V=73\times10^{-6}\times145$, we have the equation $73\times10^{-6}\times145=(73\times10^{-6}+C_2)\times18$.

Step3: Solve for $C_2$

First, expand the right - hand side: $73\times10^{-6}\times145=73\times10^{-6}\times18 + 18C_2$. Then, $73\times10^{-6}(145 - 18)=18C_2$. So, $C_2=\frac{73\times10^{-6}(145 - 18)}{18}$.
\[

$$\begin{align*} C_2&=\frac{73\times10^{-6}\times127}{18}\\ &=\frac{73\times127\times10^{-6}}{18}\\ &=\frac{9271\times10^{-6}}{18}\\ &\approx5.2\times10^{-4}\ F = 520\ \mu F \end{align*}$$

\]

Answer:

$520\ \mu F$