Capacitors

JEE Main level, numerical answer, medium

157 of 180

Question 157 of 180

A 600pF600pF capacitor is charged by 200V200V supply. It is then disconnected from the supply and is connected to another uncharged 600pF600pF capacitor. Electrostatic energy lost in the process is _____ μJ\mu J.

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Answer: 6

The formula to calculate the initial energy stored in the first capacitor is given by
Ui=12CV2...(1){U}_{i}=\dfrac{1}{2}C{V}^{2}...\left(1\right)
The formula to calculate the final energy stored when two capacitors are connected is given by
Uf=12C(V2)2×2=14CV2...(2)\begin{matrix}{U}_{f} & = & \dfrac{1}{2}C{\left(\dfrac{V}{2}\right)}^{2}\times 2 \\ & = & \dfrac{1}{4}C{V}^{2}...\left(2\right)\end{matrix}
Subtract equation (2) from equation (1) to obtain the energy loss (∆U)\left(∆U\right).
∆U=12CV2−14CV2=CV24...(3)\begin{matrix}∆U & = & \dfrac{1}{2}C{V}^{2}-\dfrac{1}{4}C{V}^{2} \\ & = & \dfrac{C{V}^{2}}{4}...\left(3\right)\end{matrix}
Substitute the values of the known parameters into equation (3) to calculate the required loss.
∆U=600×10−12×(200)24J=6μJ\begin{matrix}∆U & = & \dfrac{600\times {10}^{-12}\times (200{)}^{2}}{4}J \\ & = & 6\mu J\end{matrix}

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