Question

Each capacitor in figure (31-E21) ha a capacitance of $10\mu F.$ The emf of the battery is 100 V. Find the energy stored in each of the four capacitors.

Answer 1

$C=10\mu F=10\times {10}^{-6}F.$

For a and d

$q=4\times {10}^{-4}C,C={10}^{-5}F$

So, $E=\frac{1}{2}\times {10}^{-5}\times 40\times 40$

$=8\times {10}^{-3}J=8mJ$

For b and c

$q=4\times {10}^{-4}C,$

$C_{eq}=2C=2\times {10}^{-5}F,$

$v=20V$

So, $E=\frac{1}{2}C{v}^2$

$\frac{1}{2}\times {10}^{-5}\times 400$

$=2\times {10}^{-3}J=2mJ.$