JEE Main 2026 — Jan 22, Shift 2 — Physics solution

JEE Main2026Jan 22, Shift 2Physics

A capacitor $P$ with capacitance $10 \times 1 0^{- 6}$ F is fully charged with a potential difference of 6.0 V and disconnected from the battery. The charged capacitor $P$ is connected across another capacitor $Q$ with capacitance $20 \times 1 0^{- 6}$ F. The charge on capacitor $Q$ when equilibrium is established will be $α \times 1 0^{- 5}$ C (assume capacitor $Q$ does not have any charge initially), the value of $α$ is _____.

Your answer

Solution

Initial charge on P: $Q_{1} = C_{1} V_{1} = 10 \times 1 0^{- 6} \times 6 = 6 \times 1 0^{- 5}$ C. Q has zero charge.

When connected in parallel, charge redistributes to give a common potential:

$V = \frac{C _{1} V _{1} + C _{2} V _{2}}{C _{1} + C _{2}} = \frac{1 0 ^{- 5} \times 6 + 0}{3 \times 1 0 ^{- 5}} = 2$ V.

Charge on $Q$ : $Q_{2} = C_{2} V = 20 \times 1 0^{- 6} \times 2 = 4 \times 1 0^{- 5}$ C.

So $α = 4$ .

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