Stored energy in parallel-plate capacitor?
The problem says, "Suppose that we have a 2500-pF parallel-plate capacitor with air dielectric charged to 1400 V. The capacitor terminals are open circuited. a.) Find the stored energy, b.) If the plates are moved farther apart so that d is doubled, determine the new voltage on the capacitor, c.) Find the new stored energy."
Can someone please help?
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Energy in a Capacitor in Joules
E = ½CV² = ½(2.5e-9)(1400)²
b) Q = CV = (2.5e-9)(1400)
charge is conserved. If you double spacing, C goes to 1/2 the value.
V = Q/C = Q/(2.5e-9/2) = 2Q/(2.5e-9/2)
V = 2(2.5e-9)(1400)/(2.5e-9/2) = 2800 V
c) E = ½CV² = ½(2.5e-9/2)(2800)² = (1/4)(2.5e-9/2)(2800)²
you can do the math
hi.. Capacitance C = A Eo/d = 3 * 10^-4 * 8.80 5 x 10^-12 / 9 * 10^-3 C = 2.ninety 5 x 10^-13 Farad means saved = a million/2 CV^2 212.4 x 10^-13 Joules.. i think that's the respond.. gud success!! tc..