- V=24V
- E=3.92×10^{-4}J
- Charge=Q
- Capacitance=C
[tex]\boxed{\sf E=QV^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{E}{V^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{24^2}[/tex]
[tex]\\ \sf\longmapsto Q=\dfrac{3.92\times 10^{-4}}{576}[/tex]
[tex]\\ \sf\longmapsto Q=0.006\times 10^{-4}C[/tex]
[tex]\\ \sf\longmapsto Q=6\times 10^{-1}C[/tex]
[tex]\\ \sf\longmapsto Q=0.6C[/tex]
Now
[tex]\boxed{\sf Q=CV}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{Q}{V}[/tex]
[tex]\\ \sf\longmapsto C=\dfrac{0.6}{24}[/tex]
[tex]\\ \sf\longmapsto C=0.025F[/tex]
Note:-
- SI unit of charge is Coulomb(C)
- SI unitvof Capacitance is Farad(F)