JEE Main 2026 — Capacitance Question with Solution

Initial energy of the charged sphere is given by:

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

$U_i=\frac{1}{2}\times 100\times 10^{-12}\times (100{)}^2=5\times 10^{-7}$ J

When the identical uncharged sphere is brought in contact, the charge is shared equally and the common potential becomes:

$V^{\prime }=\frac{V}{2}=50$ V

Final total energy of the two spheres is:

$U_f=2\times \left(\frac{1}{2}C{V}^{\prime 2}\right)=C{V}^{\prime 2}$

$U_f=100\times 10^{-12}\times (50{)}^2=2.5\times 10^{-7}$ J

Change in energy is the loss of energy during charge sharing:

$\Delta U=U_i-U_f=5\times 10^{-7}-2.5\times 10^{-7}=2.5\times 10^{-7}$ J

$\Delta U=\frac{5}{2}\times 10^{-7}$ J

Comparing with $\alpha \times 10^{-7}$ J, we get $\alpha =\frac{5}{2}$.

Answer: $\frac{5}{2}$