(C) In a series combination,the charge on each capacitor is the same,so $Q_1 = Q_2$.Let the potential at point $D$ be $V$.The charge on capacitor $C_1

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$\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$

(C) In a series combination,the charge on each capacitor is the same,so $Q_1 = Q_2$.Let the potential at point $D$ be $V$.The charge on capacitor $C_1$ is $Q_1 = C_1(V_1 - V)$.The charge on capacitor $C_2$ is $Q_2 = C_2(V - V_2)$.Equating the charges: $C_1(V_1 - V) = C_2(V - V_2)$.Expanding the terms: $C_1 V_1 - C_1 V = C_2 V - C_2 V_2$.Rearranging to solve for $V$: $C_1 V_1 + C_2 V_2 = V(C_1 + C_2)$.Therefore,the potential at point $D$ is $V = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2}$.

Class 12 Physics Electrostatic Potential and Capacitance Equivalent Capacitance of Capacitor connected in Series and Parallel

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Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of point $B$ is $V_2$. The potential at point $D$ will be

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A
$\frac{1}{2}(V_1+V_2)$
B
$\frac{C_2 V_1+C_1 V_2}{C_1+C_2}$
C
$\frac{C_1 V_1+C_2 V_2}{C_1+C_2}$
D
$\frac{C_2 V_2-C_1 V_2}{C_1+C_2}$

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Electrostatic Potential and Capacitance

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Equivalent Capacitance of Capacitor connected in Series and Parallel

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Two capacitors $C_1$ and $C_2$ in a circuit are joined as shown in the figure. The potential of point $A$ is $V_1$ and that of point $B$ is $V_2$. The potential at point $D$ will be

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