Work done by moving charge from one sphere to another?

  • #1

Homework Statement


Consider two metal spheres, sphere 1 having radius R1 = 20 cm, and sphere 2 having a radius R2 = 10 cm. The two are rather close to one another, separated by a center-to-center distance of only 80 cm. Suppose now that they are connected to each other by a thin wire that is broken open in the middle by a switch. Initially sphere 1 is charged to 5 nC, and sphere 2 is neutral. When the switch is closed, charge moves spontaneously from sphere 1 to sphere 2 un- til equilibrium is reached. (For simplicity, assume that the charge is uniformly distributed on their surfaces, neglecting the fact that the two spheres tend to polarize one another, and assume that very little charge remains on the con- necting wire because it is so thin.) What is the maximum work (in Joules) that could be extracted from the discharge process if one were so inclined?


Homework Equations


Q1/R1 = Q2/R2
W = Q2/2C

The Attempt at a Solution


I calculated how much charge moved from sphere 1 to sphere 2:
Q1/R1 = Q2/R2
Q2 = 10(5 - Q2)/20
Q2 = 1.66 nC

I don't know how to convert that charge into work done, since it isn't a capacitor or a uniform electric field. If I calculate the voltage between the two spheres I could treat them as I would a capacitor, but I'm not sure how to do that. Help please?
 

Answers and Replies

  • #2
501
66
What is the initial potential energy? The final?
I calculate the voltage between the two spheres I could treat them as I would a capacitor, but I'm not sure how to do that.
What is the necessary condition for a two-body system to be a capacitor? (Think in terms of charge) Is this satisfied here?

Hope this helps.
 
  • #3
To be a capacitor there just has to be a charge difference between the two objects (voltage), but I'm not sure how to calculate the potential energy, that's the problem.
 
  • #4
gneill
Mentor
20,906
2,857
You might want to investigate the self-capacitance of a conducting sphere.
 
  • #5
I just don't understand how to figure out the capacitance, because it depends on the distance between the plates or whatever shape the capacitor is on. Voltage also depends on distance. Since the charge is traveling through a wire, distance doesn't have any affect on the problem (correct me if I'm wrong), so what do I use in place of distance to find the capacitance/voltage?
 
  • #6
Oh, ok I found this equation for the self capacitance of a sphere
2cf5e921d437ccb90f23b9dd839d25c7.png


The first few pages I looked at had other things that weren't useful.
 
  • #7
Using that equation, the work would be W = Q22/8πεoR2 = 1.248*1012 = 1.25 x 10-6 J. Is that right?
 
  • #8
I think you would actually do the calculation using R1 because it's losing energy, but shouldn't the energy gained by the second sphere equal the energy lost by the first sphere? I don't understand which capacitance to use when calculating the work. It seems like the capacitance should be a value that involves both of the spheres, but I have yet to find an equation or anything for that.
 
  • #9
gneill
Mentor
20,906
2,857
You have an initial stored energy associated with the first spherical capacitor. After the charge movement you have two capacitors, each with some charge, and so each storing some amount of energy. The total stored energy is then the sum of the two.

So you have some initial stored energy and some final stored energy...
 
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