Question

A capacitor is charged by a cell of emf $E$ and the charging battery is then removed. If an identical capacitor is now inserted in the circuit in parallel with the previous capacitor, the potential difference across the new capacitor is :

• A. $2E$
• B. $E$
• C. $E/2$
• D. zero

Answer 1

The correct option is C $E/2$

As the battery is disconnected so total is constant. i.e $Q_t=CE$
When a identical capacitor is add in parallel so the total capacitance is $C_t=C+C=2C$.
Now the common potential $=\frac{totalcharge}{totalcapacity}=\frac{CE}{2C}=\frac{E}{2}$