(B) Initial energy $U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 100 \times 10^{-12} \times (100)^2 = 0.5 \times 10^{-6} \text{ J}$.After contact, cha

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(B) Initial energy $U_i = \frac{1}{2} C V^2 = \frac{1}{2} \times 100 \times 10^{-12} \times (100)^2 = 0.5 \times 10^{-6} \text{ J}$.After contact, charge is shared equally because the spheres are identical $(C_1 = C_2 = 100 \text{ pF})$.The common potential $V' = \frac{C_1 V + C_2(0)}{C_1 + C_2} = \frac{V}{2} = 50 \text{ V}$.Total capacitance $C_{eq} = C_1 + C_2 = 200 \text{ pF}$.Final energy $U_f = \frac{1}{2} C_{eq} (V')^2 = \frac{1}{2} \times 200 \times 10^{-12} \times (50)^2 = 0.25 \times 10^{-6} \text{ J}$.Energy loss $\Delta U = U_i - U_f = 0.5 \times 10^{-6} - 0.25 \times 10^{-6} = 0.25 \times 10^{-6} \text{ J} = 2.5 \times 10^{-7} \text{ J}$.Comparing with $\alpha \times 10^{-7} \text{ J}$, we get $\alpha = 2.5 = \frac{5}{2}$.

Class 12 Physics Electric Potential and Capacitance Sharing of Charge in Capacitor Circuit

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$A$ sphere of capacitance $100 \text{ pF}$ is charged to a potential of $100 \text{ V}$. Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha \times 10^{-7} \text{ J}$. The value of $\alpha$ is . . . . . . . (combined capacitance of spheres is $200 \text{ pF}$)

JEE MAIN 2026 All JEE MAIN

A
$5$
B
$\frac{5}{2}$
C
$\frac{7}{2}$
D
$\frac{9}{2}$

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Class 12

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Electric Potential and Capacitance

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Sharing of Charge in Capacitor Circuit

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If $n$ drops, each of capacitance $C$, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

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$C_1$ धारिता वाले एक संधारित्र को $V_1$ विभव तक आवेशित किया जाता है और फिर वियोजित कर दिया जाता है। $C_2$ धारिता वाले एक अनावेशित संधारित्र को $C_1$ के साथ समांतर क्रम में जोड़ा जाता है। परिणामी विभव $V_2$ है:

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An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

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$A$ $60\; pF$ capacitor is fully charged by a $20\; V$ supply. It is then disconnected from the supply and connected to another uncharged $60\; pF$ capacitor in parallel. The electrostatic energy that is lost in this process by the time the charge is redistributed between them is (in $nJ$):

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Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on the capacitors will be:

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If $n$ drops, each of capacitance $C$, coalesce to form a single big drop, then the ratio of the energy stored in the big drop to that in each small drop will be

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel plate capacitor charged to a potential $V$. The two capacitors share charges and the common potential becomes $V^{\prime}$. The dielectric constant $K$ is

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Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially,the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state,the charge on the capacitors will be:

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