Question

When 24.0 V is applied to a

Answer 1

• V=24V
• E=3.92×10^{-4}J
• Charge=Q
• Capacitance=C

$\boxed{\sf E=QV^2}$

$\\ \sf\longmapsto Q=\dfrac{E}{V^2}$

$\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}$

$\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}$

$\\ \sf\longmapsto Q=0.006\times 10^{-4}C$

$\\ \sf\longmapsto Q=6\times 10^{-1}C$

$\\ \sf\longmapsto Q=0.6C$

Now

$\boxed{\sf Q=CV}$

$\\ \sf\longmapsto C=\dfrac{Q}{V}$

$\\ \sf\longmapsto C=\dfrac{0.6}{24}$

$\\ \sf\longmapsto C=0.025F$

Note:-

• SI unit of charge is Coulomb(C)
• SI unitvof Capacitance is Farad(F)