Question

When 24.0 V is applied to a

capacitor, it stores 3.92 x 10-4 J of
energy. What is the capacitance?
[?] x 10!? E

Answer 1

Answer: 1.36 x 10^-6

Explanation: acellus

Answer 2

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

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

`\\ \sf\longmapsto Q=(E)/(V^2)`

`\\ \sf\longmapsto Q=(3.92* 10^(-4))/(24^2)`

`\\ \sf\longmapsto Q=(3.92* 10^(-4))/(576)`

`\\ \sf\longmapsto Q=0.006* 10^(-4)C`

`\\ \sf\longmapsto Q=6* 10^(-1)C`

`\\ \sf\longmapsto Q=0.6C`

Now

`\boxed{\sf Q=CV}`

`\\ \sf\longmapsto C=(Q)/(V)`

`\\ \sf\longmapsto C=(0.6)/(24)`

`\\ \sf\longmapsto C=0.025F`

Note:-

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