How Is Energy Calculated in a Charged Capacitor?

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Homework Statement

1) A 4.7-mu or micro FF capacitor is charged to 290 V. How much energy is stored in the capacitor?

2) How much additional energy is required to charge the capacitor from 290 V to 780 V?

Homework Equations

U = 1.2*C*V²

The Attempt at a Solution

The first one I got right...

1) U = 1/2*(4.7 * 10^-6)*290 = 1.975E-1

2) I thought to find this one you could use the change in energy would have to equal the change in potential? So I tried this...but its not right, can anyone tell me where I went wrong? thanks!

Uf - 1.975E-1 = .5*(4.7E-6)(780)² - .5*4.7E-6*290²

Uf = 1.4297 J

 

[cse63146]

I did Uf - 1.975E-1 = .5*(4.7E-6)(780)2 - .5*4.7E-6*290^2

but I got 1.232

 

[nlightner]

To answer this question, we need to use the formula for energy stored in a capacitor, which is U = 1/2*C*V^2.

1) Plugging in the given values, we get U = 1/2*(4.7*10^-6)*290^2 = 1.975 J. Therefore, the energy stored in the capacitor is 1.975 J.

2) To find the additional energy required to charge the capacitor from 290 V to 780 V, we need to find the difference in energy between the final and initial states. So, ΔU = Ufinal - Uinitial = 1/2*C*(780^2 - 290^2) = 1.429 J. Therefore, the additional energy required is 1.429 J.

The mistake in your attempt was that you only considered the change in potential, but you also need to consider the change in capacitance. Hope this helps!