Consider two uncharged capacitors of equal capacitance 200 pF. One of them is charged by a 100 V supply and disconnected. Now this capacitor is connected to the uncharged capacitor. The amount of electrostatic energy lost in the process is:

Question

Accepted Answer

A. 1. Recall the standard result for the **energy lost when a charged capacitor is connected to an identical uncharged capacitor**.
When two capacitors $C_1$ and $C_2$ share charge, the energy lost is given by:
$$U_{loss}=\frac{1}{2}\cdot\frac{C_1C_2}{C_1+C_2}\cdot V^2$$

2. Substitute the given values, $C_1=C_2=200\ pF=200	imes10^{-12}F$ and $V=100\ V$.
$$U_{loss}=\frac{1}{2}\left(\frac{200	imes200}{200+200}ight)	imes10^{-12}	imes(100)^2$$

3. Simplify the coefficient inside the bracket first.
$$\frac{200	imes200}{400}=100$$

4. Substitute this back and multiply through to get the final energy loss.
$$U_{loss}=\frac{1}{2}	imes100	imes10^{-12}	imes10^4=\frac{1}{2}	imes10^{-6}=0.5	imes10^{-6}\ J$$

Hence, the electrostatic energy lost is $0.5	imes10^{-6}\ J$, matching **Option A**.