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Q. A capacitor of capacitance $C$ is charged with the help of a $200\, V$ battery. It is then discharged through a small coil of resistance wire embedded in a thermally insulated block of specific heat capacity $2.5 \times 10^{2} \,J / kg$ and mass $0.1 \,kg$. If the temperature of the block rises by $0.4\, K$, the value of $C$ is

Electrostatic Potential and Capacitance

Solution:

$\frac{1}{2} \times C \times(200)^{2}$
$=2.5 \times 10^{2} \times 0.1 \times 0.4$
$2 \times 10^{4} C=1 \times 10$
$C=\frac{1}{2 \times 10^{3}}=500 \,\mu F$