Q.
The energy stored in the capacitor as shown in the figure (a) is $4.5 \times 10^{- 6}J.$ If the battery is replaced by another capacitor of 900 $pF$ as shown in figure (b), then the total energy of system is
A) $4.5 \times 10^{- 6}\ J$
B) $2.25 \times 10^{- 6}\ J$
C) Zero
D) $9 \times 10^{- 6}\ J$
Explanation
(b) Energy stored in the capacitor $= \frac{1}{2}\frac{Q^{2}}{C}$
$Q = CV = 900 \times 10^{- 12}F \times 100V$
$\therefore Q = 9 \times 10^{- 8}C$
Energy of the capacitor when fully charged
$= \frac{1}{2}\frac{Q^{2}}{C} = 4.5 \times 10^{- 6}\ J$
[image:1_image403.emf]
The total charge is conserved. In figure (b), total capacitance
$= C^{''} = 2 \times C = 2 \times 900\ pF$
$\therefore$ Final energy $= \frac{1}{2}\frac{Q^{2}}{C''} = \frac{1}{2}.\frac{Q^{2}}{2C}$
$\therefore$ Final energy $= \frac{4.5 \times 10^{- 6}\ J}{2} = 2.25 \times 10^{- 6}\ J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.