24.17 • in fig. e24.17, each capacitor has (c = 4.00 mu f) and (v_{ab}=+28.0 v). calculate (a) the charge on…

Explanation:

Step1: Calculate equivalent capacitance of $C_1$ and $C_2$

$C_{12}=C_1 + C_2$ (capacitors in parallel), since $C_1 = C_2=4.00\ \mu F$, then $C_{12}=4.00\ \mu F+4.00\ \mu F = 8.00\ \mu F$

Step2: Calculate equivalent capacitance of $C_{12}$ and $C_3$

$C_{123}=\frac{C_{12}\times C_3}{C_{12}+C_3}$ (capacitors in series), with $C_{12} = 8.00\ \mu F$ and $C_3 = 4.00\ \mu F$, so $C_{123}=\frac{8.00\ \mu F\times4.00\ \mu F}{8.00\ \mu F + 4.00\ \mu F}=\frac{32}{12}\ \mu F=\frac{8}{3}\ \mu F$

Step3: Calculate equivalent capacitance of $C_{123}$ and $C_4$

$C_{eq}=C_{123}+C_4$ (capacitors in parallel), $C_{123}=\frac{8}{3}\ \mu F$ and $C_4 = 4.00\ \mu F=\frac{12}{3}\ \mu F$, then $C_{eq}=\frac{8 + 12}{3}\ \mu F=\frac{20}{3}\ \mu F$

Step4: Calculate the total charge $Q_{total}$

$Q_{total}=C_{eq}V_{ab}$, with $C_{eq}=\frac{20}{3}\ \mu F$ and $V_{ab}=28.0\ V$, so $Q_{total}=\frac{20}{3}\times10^{- 6}\ F\times28.0\ V=\frac{560}{3}\times10^{-6}\ C$

Step5: Calculate the potential difference across $C_4$

$V_4=\frac{Q_4}{C_4}$, and since $Q_4 = Q_{total}$ (in parallel - branch with $C_4$), $V_4=\frac{\frac{560}{3}\times10^{-6}\ C}{4\times10^{-6}\ F}=\frac{140}{3}\ V$

Step6: Calculate the potential difference across $C_{123}$

$V_{123}=V_{ab}-V_4=28.0\ V-\frac{140}{3}\ V=\frac{84 - 140}{3}\ V=-\frac{56}{3}\ V$ (magnitude is $\frac{56}{3}\ V$)

Step7: Calculate the charge on $C_3$

$Q_3 = C_3V_{123}$, $C_3 = 4\times10^{-6}\ F$ and $V_{123}=\frac{56}{3}\ V$, so $Q_3=4\times10^{-6}\ F\times\frac{56}{3}\ V=\frac{224}{3}\times10^{-6}\ C$

Step8: Calculate the charge on $C_1$ and $C_2$

Since $C_1$ and $C_2$ are in parallel and $V_{12}=V_{123}$, $Q_1 = Q_2=\frac{Q_3}{2}$ (because $C_1 = C_2$), $Q_1=Q_2=\frac{112}{3}\times10^{-6}\ C$

Step9: Calculate the potential difference across each capacitor

$V_1 = V_2=\frac{Q_1}{C_1}=\frac{\frac{112}{3}\times10^{-6}\ C}{4\times10^{-6}\ F}=\frac{28}{3}\ V$ $V_3=\frac{Q_3}{C_3}=\frac{\frac{224}{3}\times10^{-6}\ C}{4\times10^{-6}\ F}=\frac{56}{3}\ V$ $V_4=\frac{140}{3}\ V$

Step10: Calculate the potential difference between $a$ and $d$

$V_{ad}=V_{123}=\frac{56}{3}\ V\approx18.7\ V$

Answer:

(a) $Q_1 = Q_2=\frac{112}{3}\times10^{-6}\ C\approx37.3\times10^{-6}\ C$, $Q_3=\frac{224}{3}\times10^{-6}\ C\approx74.7\times10^{-6}\ C$, $Q_4=\frac{560}{3}\times10^{-6}\ C\approx186.7\times10^{-6}\ C$ (b) $V_1 = V_2=\frac{28}{3}\ V\approx9.33\ V$, $V_3=\frac{56}{3}\ V\approx18.7\ V$, $V_4=\frac{140}{3}\ V\approx46.7\ V$ (c) $V_{ad}=\frac{56}{3}\ V\approx18.7\ V$