Question

problem 17.46
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 ($c_1$) is then connected (see (figure 1) b) to a second (initially uncharged) capacitor, $c_2$, the final voltage on each capacitor is 18 v.
part a
what is the value of $c_2$? hint: charge is conserved.
express your answer using two significant figures and include the appropriate units.

Explanation:

Step1: Calculate initial charge of $C_1$

The formula for charge on a capacitor is $Q = C V$. Initially, $C_1=7.3\ \mu F$ and $V = 145\ V$. So, $Q_1 = C_1V=(7.3\times10^{- 6}\ F)\times145\ V = 1.0585\times10^{-3}\ C$.

Step2: Calculate total capacitance after connection

After connection, the voltage across both capacitors is $V_f = 18\ V$. The total charge $Q_{total}=Q_1$ (charge - conservation). Using $Q = C_{total}V_f$, we can find $C_{total}=\frac{Q_1}{V_f}=\frac{1.0585\times10^{-3}\ C}{18\ V}\approx58.8\times10^{-6}\ F$.

Step3: Find $C_2$

Since $C_{total}=C_1 + C_2$, then $C_2=C_{total}-C_1$. Substituting $C_1 = 7.3\times10^{-6}\ F$ and $C_{total}\approx58.8\times10^{-6}\ F$, we get $C_2=(58.8 - 7.3)\times10^{-6}\ F = 51.5\times10^{-6}\ F\approx52\ \mu F$.

Answer:

$52\ \mu F$