Energy stored in parallel-plate capacitor with changing plate separation

[gunitsoldier9]

Homework Statement

A parallel-plate capacitor has plates with an area of 410 cm^2 and an air-filled gap between the plates that is 2.00 mm thick. The capacitor is charged by a battery to 560 V and then is disconnected from the battery.
a.How much energy is stored in the capacitor?
b.The separation between the plates is now increased to 4.10 mm. How much energy is stored in the capacitor now?
c.How much work is required to increase the separation of the plates from 2.00 mm to 4.10 mm?

Homework Equations

C=(E₀A/d)
U=1/2CV²

The Attempt at a Solution

I got part a but it keeps saying that part b is wrong when i do it the same way. I combined the first 2 eqns
U= 1/2(E₀A/d)V² for part A. Help please

 

[quantumdude]

Post what you've done, then we can check it.

 

[gunitsoldier9]

U= 1/2(E₀A/d)V²=(1/2)(8.85*10^-12*.041m/.002m)*560²=28.4 microjoules

Same thing for part b but with the new the new d
U= 1/2(E₀A/d)V²=(1/2)(8.85*10^-12*.041m/.0041m)*560² = 13.9 microjoules but its wrong

 

[quantumdude]

I got what you got. Perhaps the answer key has a typo?

 

[mplayer]

Maybe the answer in the back of the book was accidentally figured as "separation between the plates is now increased BY 4.10 mm" instead of "separation between the plates is now increased TO 4.10 mm".

Try computing 'U' with d = (0.002m + 0.0041m).

 

[gunitsoldier9]

Maybe the answer in the back of the book was accidentally figured as "separation between the plates is now increased BY 4.10 mm" instead of "separation between the plates is now increased TO 4.10 mm".

Try computing 'U' with d = (0.002m + 0.0041m).

Nope, no luck. Beginning to think there is a mistake with the problem

 

[uvaj23]

I was having the same problem, and I hope that you still need help with this. But for any people in the future who need help, please refer here: http://www.physics.miami.edu/~korotkova/PHY102_Lecture4_09.pdf" . Specifically on slide 8.

If for some reason that doesn't work, just read:

multiply answer (a) by the factor that your distance increased. For instance, if your distance doubled, multiply (a) by two. For part c, you simply take the difference of (b) and (a).