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Electric Capacitance, Energy of an Electric Field

Problem 3.123

In the circuit shown in Fig. 3.25 the emf of each battery is equal to E=60 V,\mathscr{E}=60 \mathrm{~V}, and the capacitor capacitances are equal to C1=2.0μFC_{1}=2.0 \mu \mathrm{F} and C2=3.0μF.C_{2}=3.0 \mu \mathrm{F} . Find the charges which will flow after the shorting of the switch SwS w through sections 1,2 and 3 in the directions indicated by the arrows.

Reveal Answer
 3.123. q1=EC1(C1−C2)/(C1+C2)=−24μCq2=EC2(C1−C2)/(C1+C2)=−36μC,q3=E(C2−C1)=+60μC\begin{aligned} &\text { 3.123. } q_{1}=\mathscr{E} C_{1}\left(C_{1}-C_{2}\right) /\left(C_{1}+C_{2}\right)=-24 \mu \mathrm{C}\\ &q_{2}=\mathscr{E} C_{2}\left(C_{1}-C_{2}\right) /\left(C_{1}+C_{2}\right)=-36 \mu \mathrm{C}, q_{3}=\mathscr{E}\left(C_{2}-C_{1}\right)=+60 \mu \mathrm{C} \end{aligned}
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