MHT CET202613 April 2026Morning ShiftPhysicsCapacitanceActual
Two identical parallel plate air capacitors are connected in series to a battery of e.m.f. 'V'. If one of the capacitor is inserted in liquid of dielectric constant 'K' then potential difference of the other capacitor will become
Options
- AKV K+1
- BKV K-1
- CK+1 KV
- DK-1 KV
Correct answer
A. KV K+1
Step-by-step solution
Let the capacitance of each identical parallel plate air capacitor be C . When one capacitor is inserted in a liquid of dielectric constant K , its capacitance becomes C₁ = KC . The capacitance of the other capacitor remains C₂ = C . Since the two capacitors are connected in series across the battery of e.m.f. V , the equivalent capacitance of the circuit is C_ eq = C₁ C₂ C₁ + C₂ = KC C KC + C = KC K+1 . The total charge supplied by the battery is Q = C_ eq V = KCV K+1 . In a series combination, the charge on each