Question 586: Electrostats & Capacitance

Question

Two identical capacitors have the same capacitance '$$\mathrm{C}$$'. One of them is charged to potential '$$\mathrm{V_1}$$' and the other to $$\mathrm{V_2}$$. The negative ends of the capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

Accepted Answer

$$\begin{aligned}
& Q_1=C V_1, Q_2=C V_2, Q=Q_1+Q_2=2 C V \
& 	herefore C V_1+C V_2=2 C V \
& 	herefore \quad V=\frac{V_1+V_2}{2}
\end{aligned}$$
$$\begin{aligned}
&	ext { Decrease in energy }  =\frac{1}{2} C V_1^2+\frac{1}{2} C V_2^2-\frac{1}{2}(2 C)\left(\frac{V_1+V_2}{2}ight)^2 \
& =\frac{1}{2} C\left[V_1^2+V_2^2-2\left(\frac{V_1+V_2}{2}ight)^2ight] \
& =\frac{1}{2} C\left[V_1^2+V_2^2-\frac{1}{2}\left(V_1+V_2ight)^2ight] \
& =\frac{1}{4} C\left[2 V_1^2+2 V_2^2-\left(V_1+V_2ight)^2ight] \
& =\frac{1}{4} C\left(V_1-V_2ight)^2
\end{aligned}$$

Answer

$$\frac{1}{4} C\left(V_1-V_2\right)^2$$

Answer

$$\frac{1}{2} C\left(v_1^2+v_2^2\right)$$

Answer

$$\frac{1}{2} C\left(V_1^2-V_2^2\right)$$

Answer

$$\frac{1}{2} C\left(V_1+V_2\right)^2$$

Electrostats & Capacitance

Two identical capacitors have the same capacitance '$\mathrm{C}'. One of them is charged to potential '\mathrm{V_1}' and the other to \mathrm{V_2}$. The negative ends of the capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

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