Finding charge on sphere using energy

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Homework Statement


a) What charge would need to be placed on a metal sphere such that the electric field produced stored a total of 50 J if the radius of the sphere was 18 cm?
b) How much work would need to be done to reduce the radius to 9 cm, assuming the sphere is (mechanically) easily compressible?

Homework Equations


V=(4/3)pi r2

The Attempt at a Solution


Not sure where to tackle this problem...
 
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How much work to get the first smallest bit of charge onto the sphere (from infinity)?
How about the next bit?
The bit after that?
... spot the pattern.

Either that or look up the equation for the amount of energy stored in a charged metal sphere.
 
Simon Bridge said:
How much work to get the first smallest bit of charge onto the sphere (from infinity)?
How about the next bit?
The bit after that?
... spot the pattern.

Either that or look up the equation for the amount of energy stored in a charged metal sphere.

I found this equation from the textbook.

u = 2keQ2 / R

Using this I get

50 = 2(9X109)Q2 / 0.18
0.18(50) = 2keQ2
Q2 = 9 / 2k
Q = square root (9 / 2(9X109)
Q = 2.23607 X 10-5
2.24X10-5 C

But this isn't right...
 
Simon Bridge said:
So how can you tell if it is the right equation?

i.e. read the book carefully: what exactly does it say this is the energy of?

Energy stored by a charged sphere
 
need_aca_help said:
Energy stored by a charged sphere
... is that exactly what it says?

Not just what's written down - there is usually a diagram as well.

i.e. is the sphere uniformly charged? Is the charge distributed through the whole volume or just on the surface? Is the sphere a conductor or an insulator?

If you cannot narrow this down you will have to do some calculus.
 
Simon Bridge said:
... is that exactly what it says?

Not just what's written down - there is usually a diagram as well.

i.e. is the sphere uniformly charged? Is the charge distributed through the whole volume or just on the surface? Is the sphere a conductor or an insulator?

If you cannot narrow this down you will have to do some calculus.

I managed to solve the problem, using this equation:
E = 1/2QV
E = 1/2Q(kQ/r)