# When two spheres of equal charge make contact, why does the larger sphere gain more charge?

Suppose that two spheres, $S_1$ and $S_2,$ with radii $R_1$ and $R_2$ resp. have the same uniform charge $Q$ and $R_1 > R_2.$ After they are forced to come in contact, why does $S_1$ gain more charge? Why don't they both continue to have equal charges?

• It is more energetically favorable for the larger sphere to take on the charge from the smaller – D. W. Aug 31 '14 at 9:16

A spherically symmetric charged body behaves as a point charge. By this I mean that the electric potential outside the sphere is the same as for a point charge at the centre of the sphere. This is a consequance of Gauss's law.

So suppose we have a sphere of radius $R$ and charge $Q$, the potential at a distance $r > R$ from the centre of the sphere is just:

$$V = -\frac{1}{4\pi\epsilon_0}\frac{Q}{r}$$

And the voltage at the surface of there sphere is therefore:

$$V_{surface} = -\frac{1}{4\pi\epsilon_0}\frac{Q}{R} \tag{1}$$

Now consider your two spheres of radii $R_1$ and $R_2$, and charges $Q_1$ and $Q_2$. When the spheres are touched together they will have the same potential because current can flow between them until the potential difference is zero. That means:

$$V_1 = V_2$$

and using equation (1) for the surface voltage we find:

$$-\frac{1}{4\pi\epsilon_0}\frac{Q_1}{R_1} = -\frac{1}{4\pi\epsilon_0}\frac{Q_2}{R_2}$$

or more simply:

$$\frac{Q_1}{R_1} = \frac{Q_2}{R_2}$$

And this answers your question because if we rearrange this to get $Q_1$ we get:

$$Q_1 = Q_2\frac{R_1}{R_2}$$

and since we know $R_1 > R_2$ it follows that $Q_1 > Q_2$.

The ratio of $Q/V$ is called the capacitance, or in this case it would be more precise to call it the self capacitance, and the capacitance of a conducting sphere is:

$$C = \frac{Q}{V} = 4\pi\epsilon_0 R$$

Another way of answering your question is to point out that the capacitance of a large sphere is greater than the capacitance of a small sphere, so when the voltages are equal the large sphere will contain a greater charge.

The naive reasoning which leads to the conclusion that charges $Q_1$ and $Q_2$ of two touching conducting spheres with radii $R_1$ and $R_2$ are related by the relation $Q_1 = Q_2\frac{R_1}{R_2}$ is wrong. This formula holds only when the distance between the spheres $L$ is large compared to $R1$ and $R_2$, $L\gg R_1,R_2$, and the spheres are connected by a long wire.

The complete rigorous solution to this question can be found by using some formulas given in the book "Problems in Electrodynamics" by V.V. Batygin, I.N. Toptygin http://www.amazon.com/Problems-Electrodynamics-English-Russian-Edition/dp/0120821605 , for example see problems 211 and 67. In the notation used in that book, the case of touching spheres corresponds to taking the limit $a\to 0$ in the bispherical coordinates $(\xi,\eta,\alpha)$: $$x=\frac{a\sin\eta\cos\alpha}{\cosh\xi-\cos\eta},\quad y=\frac{a\sin\eta\sin\alpha}{\cosh\xi-\cos\eta},\quad z=\frac{a\sinh\eta}{\cosh\xi-\cos\eta}$$ and solving the corresponding boundary problem. The surfaces of two almost touching spheres will have coordinates $\xi_1=\frac{a}{R_1}\to 0$,$\ \xi_2=-\frac{a}{R_2}\to 0$. Without going into all the details the final answer is $$\frac{Q_1}{Q_2}= \frac{R_1}{R_2}\frac{1+2\int\limits_0^\infty\displaystyle\frac{e^{-t}\cosh[(1+k)t]-e^{-kt}}{\sinh[(1+k)t]}e^{-t}dt}{1+2\int\limits_0^\infty\displaystyle\frac{e^{-t}\cosh[(1+1/k)t]-e^{-t/k}}{\sinh[(1+1/k)t]}e^{-t}dt},\quad \text{where}\ \ k=\frac{R_1}{R_2}.$$

But if one is interested only in qualitative answer to the question, then as it was explained in the comment by D.W. above, it is energetically more favorable for the sphere with the largest radius to absorb more charge. It is the most obvious in the case when one of the spheres has much larger radius than the other $R_1\gg R_2$.

Consider an extreme case, where the two spheres both have just a little charge, and one of them is far larger than the other. Say one is molecule-sized and the other is Earth-sized, and they each just have two excess electrons. The two electrons on the small ball are quite close, and would prefer to be farther away from each other. So, when the large ball offers them some new space to lease, one of them gladly takes it, and the two electrons already on the large ball hardly mind at all.

In slightly more precise terms: the charges on the small ball, because they're closer together, repel each other more strongly than the net charge on the large ball repels them, so some of them jump over.

## protected by rob♦Dec 15 '17 at 19:36

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