# Two conducting spheres connected by a wire

In the post The physics behind grounding a conductor, I explained how grounding a conductor works. There I considered the problem of determining the electrostatic equilibrium state that two conducting spheres reach after being connected by a wire. One reader of that post asked me how that equilibrium state is reached from the initial state. I will answer that question here. But first, let’s state more precisely the problem.

Two conducting spheres of respective radii $$a$$ and $$b$$ are connected by a wire. Before connecting the spheres, the one with radius $$a$$ is charged with a total charge of $$Q$$, while the second one is uncharged. The spheres are far apart in such a way that the electric potential of any of them is negligible near the other one. If $$R$$ is the total electric resistance of the wire, how the charge on both spheres changes with time after being connected to the wire?

Now let’s see how to answer the question.

## Physical description of the problem

At any instant $$t\geq0$$ the respective electric charges are $$Q_a(t)$$ and $$Q_b(t)$$, and the corresponding electric potentials of the spheres are

\begin{align}
\phi_a(t) &= k\frac{Q_a(t)}{a}\\
\phi_b(t) &= k\frac{Q_b(t)}{b}.
\end{align}

The voltage $$V(t)$$ between the two spheres is the difference in their electric potentials:

\begin{split}
V(t) &= \phi_a(t)-\phi_b(t) \\
&= k\frac{Q_a(t)}{a}-k\frac{Q_b(t)}{b}.
\end{split}

Conservation of the electric charge implies the relation

Q_a(t)+Q_b(t)=Q,

so that the voltage can be expressed in terms of either $$Q_a$$ only or $$Q_b$$ only. Then, for example, we can write

\begin{split}
V(t) &= k\left[\frac{Q_a(t)}{a}-\frac{Q-Q_a(t)}{b}\right]\\
&= kQ_a(t)\left(\frac{1}{a}+\frac{1}{b}\right)-k\frac{Q}{b}
\end{split}

The electric current $$I(t)$$ that flows through the wire is related to the voltage through Ohm’s law:

V(t)=I(t)R.

The electric current $$I(t)\geq0$$ is the amount of electric charge that is transferred from one sphere to the other in the unit of time. So it is equal in magnitude to the rate of decrease of the charge $$dQ_a(t)/dt$$ in the sphere of radius $a$. Notice that $$dQ_a/dt<0$$. Indeed, the charge of the sphere of radius $$a$$ is diminishing with time. Therefore, we have

I(t)=-\frac{dQ_a(t)}{dt}.

Now we can use the equations (4) and (6) in equation (5) to obtain the differential equation that describes the dynamics of $$Q_a$$:

\frac{dQ_a}{dt}+\frac{k}{R}\left(\frac{1}{a}+\frac{1}{b}\right)Q_a=\frac{kQ}{Rb}\ ;\\
Q_a(0)=Q.

The initial charge on the sphere of radius $$a$$ is assumed to be positive so that the current flows from this sphere to the other one as shown in the figure. Remember that the direction of the electric current is in the sense of motion of positive charges. Positive charges moving in one direction are equivalent to negative charges moving in the opposite direction.

Let’s see what this equation implies.

## Dependence on time of the charge

The electrostatic equilibrium is reached when there is no change of the charge with time, i.e., when $$dQ_a/dt=0$$. We see from equation (7) that the charge $$Q_a$$ in the equilibrium state is

Q_a\to \frac{Q}{1+(b/a)}

This result agrees with that obtained in the article The physics behind grounding a conductor (Compare with Eq.(4) in that article where $$a=R_1$$ and $$b=R_2$$).

The solution of the equation (7) that satisfies the initial condition is

\begin{align}
Q_a(t)&=\left(\frac{1+(b/a)e^{-t/\tau}}{1+(b/a)}\right)Q\ ;\\
\tau&=\frac{R}{k}\left(\frac{1}{a}+\frac{1}{b}\right)^{-1}.
\end{align}

We can find $$Q_b(t)$$ with help of equation (3), namely

Q_b(t)=\left(\frac{1-e^{-t/\tau}}{1+(a/b)}\right)Q\ ,

and the current $$I(t)$$ with help of equation (6):

I(t)=\frac{k}{R}\frac{Q}{a}e^{-t/\tau}.

We see that the initial current $$I(0)=kQ/(aR)$$ is consistent with equation (2) and Ohm’s law given in equation (5).

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