An 8.37 x 10^-5 F capacitor has 2.15 x 10^-4 C of charge on its plates. How much energy is stored on the capacitor?

Help me plz I am using E=1/2×Q²/C but my answer is wrong help plzz

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  • oubaas
    Lv 7
    1 year ago
    Favorite Answer

    V = Q/C

    E = C*V^2/2 = C*Q^2/(2C^2) = Q^2/2C = 2.15^2*10^-8/(8.37*2*10^-5) = 276 μ joule (2.76*10^-4)

  • 1 year ago

    E = ½CV² but Q = CV so E = ½QV but V=Q/C so E = ½Q²/C

    E = ½(2.15e-4)²/(8.37e-5) = 276µJ

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