Two spherical conductors each of capacity C are charged to potentials V and - V . These are then con

English

Gujarati

Hindi

2. Electric Potential and Capacitance

hard

Two spherical conductors each of capacity $C$ are charged to potentials $V$ and $ - V$. These are then connected by means of a fine wire. The loss of energy will be

A

Zero

B

$\frac{1}{2}C{V^2}$

C

$C{V^2}$

D

$2C{V^2}$

Solution

(c) $\Delta V = \frac{1}{2}\frac{{C \times C}}{{(C + C)}}\,|V – ( – V){|^2} = C{V^2}$

Standard 12

Physics

Share

Similar Questions

In a charged capacitor, the energy resides

easy

Two small spheres each carrying a charge $q$ are placed $r$ metre apart. If one of the spheres is taken around the other one in a circular path of radius $r$, the work done will be equal to

easy

A parallel plate capacitor of capacity ${C_0}$ is charged to a potential ${V_0}$

$(i)$ The energy stored in the capacitor when the battery is disconnected and the separation is doubled ${E_1}$

$(ii)$ The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is ${E_2}.$

Then ${E_1}/{E_2}$ value is

medium

A capacitor of $2\,\, \mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ……$\%$

hard

(NEET-2016)

A capacitor of capacitance $C$ is initially charged to a potential difference of $V$ $volt$. Now it is connected to a battery of $2V$ with opposite polarity. The ratio of heat generated to the final energy stored in the capacitor will be

hard