# Energy stored in a device (Charged Capacitor)

## Homework Statement

The metallic sphere on top of a large van de Graaf generator has a radius of 2.0 m. Suppose that the sphere carries a charge of 3x10-5 C. How much energy is stored in this device?

## Homework Equations

So the equation I must use is this: U = 1/2C(ΔV)^2
Q is given; 3 x 10 ^ -5C
To find ΔV, I'd first the formula :V = Q/(4∏ε0r)

## The Attempt at a Solution

Then it's simply a matter of plugging the numbers. The formula I'm not sure is how to find ΔV from the information, I'm not sure I'd be using the right one.

CWatters
Homework Helper
Gold Member
What's the magic word beginning with d ?

What's the magic word beginning with d ?

Oh, I have to find the charge density don't I? Bear with me, I'm still quite confused with all of that...

So, charge density of a sphere: p = Q/V, V = (4∏r3)/3

Would I use this as my Q in the above formula?

CWatters
Homework Helper
Gold Member
I meant d for differentiate .

Somewhat informally..

Q=VC
V=Q/C
dV = dQ/C

dQ = the amount of charge when fully charged - the amount of charge when discharged

= 3x10-5 - 0

I meant d for differentiate .

Somewhat informally..

Q=VC
V=Q/C
dV = dQ/C

dQ = the amount of charge when fully charged - the amount of charge when discharged

= 3x10-5 - 0

Ok I'm sorry but I'm not following you on this one, this isn't something I've seen in class

CWatters
Homework Helper
Gold Member
It's probably simpler than you think.

The max energy case is with it charged to some voltage V, The minimium energy state is with it discharged (zero volts) so ΔV is just V. Likewise for ΔQ.

U = 1/2C(V)2

now
Q=VC so
V = Q/C

then substitute for V giving

U = 1/2*C*(Q/C)2
= Q2/2C

It's probably simpler than you think.

The max energy case is with it charged to some voltage V, The minimium energy state is with it discharged (zero volts) so ΔV is just V. Likewise for ΔQ.

U = 1/2C(V)2

now
Q=VC so
V = Q/C

then substitute for V giving

U = 1/2*C*(Q/C)2
= Q2/2C

Thanks! It is simpler than I thought... So in that case I don't need the radius of the generator to solve it, it's basically an unnecessary information given?