A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?

+2 votes

(a) A 2.00-µF capacitor is connected to a 18.0-V battery. How much energy is stored in the capacitor?
µJ

(b) Had the capacitor been connected to a 6.00-V battery, how much energy would have been stored?
µJ

asked Jan 27, 2012 in Physics by apush ~Expert~ (552 points)
    

1 Answer

+1 vote
 
Best answer

The formula for the potential energy of capacitor is:

U = (1/2)*C*V^2  (where U = potential energy, C = capacitance, V = voltage)

values of these are given in the question as:

V = 18 V

C = 2*10^-6 F

plugging into the formula we get:

 

U = (1/2)*(2*10^-6 F)*(18 V)^2

U = 3.24*10^-4 J    =         324 µJ

 

Part B, just replace the voltage with 6V

and get

 

U = (1/2)*(2*10^-6 F)*(6 V)^2

U = 3.6*10^-5 J   =    36 µJ

answered Jan 28, 2012 by awesome ~Expert~ (1,479 points)
selected Jan 28, 2012 by apush



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