Calculating Energy From Charged Particles: 0.5869 J

[Istiak]

Homework Statement

Two solid conducting sphere of radius $10 cm$ and $20 cm$ having $5\mu C$ and $30\mu C$ charges respectively are kept at 2 meter distance. Then, they are connected by a conducting wire. Find the amount of heat generated in this process.

Relevant Equations

$U=\frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{r}$

Generally, energy is $U=9\times 10^{9} \times \frac{5\times 10^{-6}30\times 10^{-6}}{2+(10+20)\times 10^{-2}}=0.5869 J$

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After touching, they have charges

$q_1 and q_2 = 35\mu C-q_1$
$\frac{q_1}{10}=\frac{35\mu C-q_1}{20}$

I was wondering where 1/10 and 1/20 coefficients come from. I was thinking to convert q_2 to q_1 but end up with nothing.

 

[BvU]

Generally

That's not very general at all. Seems more like an evaluation of $U$ in the relevant equation!

And $U$ is not the only energy in this exercise ! How much energy is needed to charge each of the two spheres ?

Also: after touching, what happens and when does it stop ?

$\$

 

[Istiak]

after touching, what happens and when does it stop ?

Since they have same charges (not magnitude) so they distract. They simply stop as further as they move, at infinity? E=0 at infinity.

How much energy is needed to charge each of the two spheres ?

dunno
$U=\int \vec F \cdot d\vec r$ ?

 

[Steve4Physics]

I’ll add a few thoughts in addition to what @BvU has said.

Each sphere acts as a capacitor in its own right. The capacitance of an isolated sphere is as shown here: http://230nsc1.phy-astr.gsu.edu/hbase/electric/capsph.html#c2. After connecting, the 2 spheres are at the same potential. You may now be able to work out where ‘the 1/10 and 1/20 coefficients' come from!

Other thoughts…

Working would be clearer if you defined and used different symbols for initial and final charges, e.g. q for initial and Q for final.

The charge distributions on each sphere won’t be spherically symmetric because of the effect of the other charged sphere. However, 2m is probably large enough (relative to the spheres’ radii) for you to ignore this. (If you don’t ignore it, the problem is much harder.)

Is 2m the centre-to-centre distance or the surface-to-surface distance? The question is ambiguous but your working assumes surface-to-surface. $2+(10+20)\times^{-2}$ is wrong but we know what you mean. However, $2+0.1+0.2$ would have been easier/neater.

It is interesting that the energy-loss occurs even if the spheres and connecting wires are perfect conductors (superconductors). So ohmic-heating isn’t the key cause of the energy-loss. An interesting question is: how is the energy lost?

 

[BvU]

Since they have same charges

They start with different charges and, more important, different potentials. A conducting wire let's charges move until there is no more potential difference.

So your job in this exercise is to compute the energy of the initial configuration, to find the charge distribution at zero potential difference and the energy of that configuration. The difference must have been dissipated.

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/capsph.html#c2

Personally, I would put the sphere centers 2 m apart, but 2.15 m might also be intended. 2.30 m definitely not.

$\$

 

[Delta2]

It is interesting that the energy-loss occurs even if the spheres and connecting wires are perfect conductors (superconductors). So ohmic-heating isn’t the key cause of the energy-loss. An interesting question is: how is the energy lost?

I think we are forced to assume that there are ohmic losses, otherwise the lost energy would have to go into kinetic energy of the charges that flow through the wire, but these charges aren't free to move around, they start from the sphere of higher potential and stop at the sphere of lower potential, this stopping of the charges is what forces us to think that there are ohmic losses that make the charges stop.

 

[Steve4Physics]

I think we are forced to assume that there are ohmic losses, otherwise the lost energy would have to go into kinetic energy of the charges that flow through the wire, but these charges aren't free to move around, they start from the sphere of higher potential and stop at the sphere of lower potential, this stopping of the charges is what forces us to think that there are ohmic losses that make the charges stop.

One of my favourite demo's when introducing capacitors was to charge a high operating-voltage capacitor to 3 or 4 kV and then 'short' it.

(I didn’t use gloves or safety glasses either!)

The main energy release occurs from the spark across the small gap, the instant before the circuit completes.

There are slightly different views on the nature of the 'spark'. I believe it is primarily produced by a tiny amount of plasma (air and maybe metal ions) created by the electrons accelerating in the strong electric field. But, even in a vacuum, the electrons will lose energy radiating when accelerating across the (unavoidable) gap.

Back in the old days, school physics exam’ questions would often ask the student to calculate the energy-loss for a discharging capacitor. As a follow-up 1-mark question, the student would then be asked where the lost energy had gone. The ‘spark’ was the answer.

In a real-world situation, the energy-loss is a mixture of Ohmic losses and energy radiated from the spark.

 

[hutchphd]

I think we are forced to assume that there are ohmic losses, otherwise the lost energy would have to go into kinetic energy of the charges that flow through the wire, but these charges aren't free to move around, they start from the sphere of higher potential and stop at the sphere of lower potential, this stopping of the charges is what forces us to think that there are ohmic losses that make the charges stop.

In a real-world situation, the energy-loss is a mixture of Ohmic losses and energy radiated from the spark.

"Real World" detaIL: If these were truly perfect conductors they would still have inductace no matter what. So the sudden connection would in fact not necessarilly cause a sudden ohmic loss but rather an oscillation. Any resistance is present would cause a decaying oscillation. There would also be some radiation. So for any usual array the oscillation is damped and usually overdamped (with no discenable oscillation).