A capacitor is charged by connecting a battery across its plates. It stores energy U. Now the battery is disconnected and another identical capacior is connected across it, then the energy stores by both capacitors of the system will be

Options

A. U
B. `(U)/(2)`
C. 2U
D. `(3)/(2)U`

Answer

B

Explanation

On removing the battery after charging, the stored in the capacitor remains constant. When a capacitor is charged by connecting a across its plates, the initial energy stored.
`U= (q^(2))/(2C)`
When the battery is disconnected, then the remains constant i.e., q= constant. Now another identical capacitor is connected it i.e., the capacitors are connected in parallel, equivalent capacitance
`C_(eq) = C_(1) + C_(2)= C+C= 2C`
Thus, final energy stored by the system of capacitor
`U.= (q^(2))/(2C_(eq))`
`=(q^(2))/(2(2C))=(1)/(2)U`
`U.= (U)/(2)`