Charge Transfer from sphere to sphere

[Rose Garden]

Homework Statement

There is a metal sphere charged to 6nC. A wire then connects this sphere to another sphere which has 2x the diameter and neutral. What are the final charges on each sphere?

Homework Equations

The Attempt at a Solution

I think the idea behind this is that the charge will flow from the charged sphere to the neutral one until the neutral one has gained enough charge that there's no longer a voltage difference. However I don't know how to get voltage difference in this situation.
Instead, I solved using the idea that the final surface charge distributions of both spheres should be the same.

So I did something like this,

Q₁/r²=Q₂/(2r)² (the 4pi cancels out)
from this I get the charge on the larger ball has to be 4x that of the smaller ball, however in my textbook it says the larger ball has 2x the charge.

 

[ehild]

Write up the potential at both surfaces with respect to infinity. They have to be equal, otherwise charge will flow from the higher potential to the lower one. If the spheres are far enough the surface charge distribution is homogeneous. The potential of a sphere is the same as that or a point charge in the centre.

ehild

 

[SammyS]

What is the electric potential at the surface of a sphere of radius, R, and charge Q?

$$V(R) = k\frac{Q}{R}$$

Yes, the potential difference is zero.

$$V(R)-V(2R)=0$$

Use a charge of Q₁on one sphere, Q₂ on the other. Q₁+Q₂ = Q_Total.

 

[Rose Garden]

ok thanks guys

 

[rwisz]

Your approach is correct in solving for the final charge on each sphere. The equation you used, Q1/r2 = Q2/(2r)2, is known as the charge distribution equation and it is used to calculate the charge on a sphere based on its radius and the charge on another sphere with a known radius and charge.

To address the discrepancy between your calculation and that of your textbook, it is important to note that the charge on a sphere is directly proportional to its radius. This means that if the radius is doubled, the charge will also double. In this case, the larger sphere has twice the diameter (and therefore four times the radius) of the smaller sphere. This means that the charge on the larger sphere should also be four times that of the smaller sphere, which is consistent with your calculation.

In summary, your approach to solving this problem is correct and your calculation for the final charges on each sphere is also correct. The discrepancy with your textbook may be due to a misunderstanding of the relationship between charge and radius for a sphere.