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Question

# In Fig. we saw earlier, $C_1=$ $6.00 \mu \mathrm{F}, \quad C_2=3.00 \mu \mathrm{F}$. and $C_3=5.00 \mu \mathrm{F}$. The capacitor network is connected to an applied potential $V_{a b}$. After the charges on the capacitors have reached their final values, the charge on $C_2$ is $30.0 \mu \mathrm{C}$. (b) What is the applied voltage $V_{a b}$ ?

Solution

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(b) Now we want to determine the applied voltage across $ab$. It is given by equation 24.1 in the next form

$V_{ab} = \dfrac{Q}{C_{\text{eq}}} \tag{2}$

Where $Q$ represents the charge on the third capacitor $Q = 90.0 \mathrm{~\mu C}$ and $C_{\text{eq}}$ is the equivalent capacitance on the system (between $C_{3}$ and ${C}'_{\text{eq}}$) in series. In series the equivalent capacitance $C_{\text{eq}}$ could be calculated by equation 24.5 in the next form

$\begin{gathered} \dfrac{1}{C_{\text{eq}}} = \dfrac{1}{{C}'_{\text{eq}}} + \dfrac{1}{C_{3}} \tag{3} \end{gathered}$

Where ${C}'_{\text{eq}}$ represents the equivalent capacitance between $C_{1}$ and $C_{2}$ and as both capacitors are in parallel, hence ${C}'_{\text{eq}}$ is given by equation 24.7

${C}'_{\text{eq}} = C_{1} + C_{2}$

Now let us plug our expression for ${C}'_{\text{eq}}$ into equation (3) to find $C_{\text{eq}}$

\begin{aligned} \dfrac{1}{C_{\text{eq}}} &= \dfrac{1}{C_{1} + C_{2}} + \dfrac{1}{C_{3}} \\ \dfrac{1}{C_{\text{eq}}} &= \dfrac{1}{6.00 \mathrm{~\mu F} + 3.00 \mathrm{~\mu F}} + \dfrac{1}{5.00 \mathrm{~\mu F}}\\ \dfrac{1}{C_{\text{eq}}} &= 0.311 \tag{Reciprocal}\\ C_{\text{eq}} &= 3.21 \mathrm{~\mu F} \end{aligned}

Now we can plug our values for $C_{\text{eq}}$ and $Q$ into equation (2) to get $V_{ab}$

\begin{aligned} V_{ab} = \dfrac{Q}{C_{\text{eq}}} = \dfrac{90.0 \mathrm{~\mu C}}{3.21 \mathrm{~\mu F}} = \boxed{28.0 \,\text{V}} \end{aligned}

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