JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Capacitance

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 ______.

Answer: 4

Solution

The charge initially stored on

P

Q = C_{P} V = 10 \times 1 0^{- 6} \times 6 = 6 \times 1 0^{- 5} C

. When

P

is connected across the uncharged

Q

, charge redistributes until both capacitors reach a common potential

V^{'} = \frac{Q}{C _{P} + C _{Q}} = \frac{6 \times 1 0 ^{- 5}}{( 10 + 20 ) \times 1 0 ^{- 6}} = 2 V .

The charge that settles on

Q

Q_{Q} = C_{Q} V^{'} = 20 \times 1 0^{- 6} \times 2 = 4 \times 1 0^{- 5} C

. Hence

α = 4

JEE Main · 2026 · 22 Jan 2026 · Shift 2 · Integer answer