Two capacitor of capacitance `2muF and 3muF` respectively are charged to potential difference 20 V each. Now the capacitors are connected such that the positively chargd plate of one capacitory is connected to the negatively charged plate of the other. after the current in the circuit has become negligible, the potential difference across the `2muF` capacitor is ____V.

Answer

4

Explanation

Initially, charge on the `2muF` capacitor `=(2)(20)=40muC`, charge on the `3muF` capacitor `=(3)(20)=60muC`
Let the final charges on the two capacitors be `Q_(1) and Q_(2)` respectively.
When steady state is achieved, the polarity of the capacitor that initially had a lower amount of charge, i.e. the `2muF` capacitor, has reversed, and the potential difference across the capacitors is equal so,
`(Q_(1))/(2)=(Q_(2))/(3)`
Now, we can choose a pair of connected plates and conserve the total charge on them as they are an isolated system.
`Q_(1)+Q_(2)=60+(-40)`
Solving, we get `Q_(1)=8muC and Q_(2)=12muC`.
so, potential differenec across the `2muF` capacitor `=(8)/(2)4V`.