A conducting sphere S1 of radius r is attached to an insulating handle. Another conducting sphere S2

Q: Why do the charges distribute in the ratio of their radii?

Because when two conductors touch, they reach an equipotential state ($V_1 = V_2$). Since $V = \frac{q}{4\pi\varepsilon_0 R}$, equating the potentials gives $\frac{q_1}{r} = \frac{q_2}{R}$, meaning charge is directly proportional to radius.

Q: Why doesn't $S_2$ just take all the charge?

Charge sharing is driven by potential difference. As charge flows to $S_2$, its potential rises, and $S_1$'s potential drops. The flow stops exactly when their potentials balance out, which always leaves some charge on $S_1$.

Q: Is the total charge conserved in each step?

Yes! During the brief moment of contact, the total charge of the system ($Q$ from $S_1$ plus whatever is already on $S_2$) is perfectly conserved and just redistributed.

Q: Why does the charge on $S_2$ reach a maximum limit?

Every time $S_1$ touches $S_2$, it brings a fresh charge $Q$, but it also takes away a fraction of the total charge. Eventually, the amount of charge $S_1$ brings is exactly equal to the amount it takes away, creating a dynamic equilibrium.

Q: What happens to the energy lost during these contacts?

Whenever charge flows between conductors at different potentials, energy is dissipated as heat (and a tiny bit of electromagnetic radiation) in the connecting wires or at the point of contact. The total energy is not conserved, only the charge is!

Question

A conducting sphere $S_1$ of radius $r$ is attached to an insulating handle. Another conducting sphere $S_2$ of radius $R$ is mounted on an insulating stand. $S_2$ is initially uncharged.
$S_1$ is given a charge $Q$, brought into contact with $S_2$ and removed. $S_1$ is recharged such that the charge on it is again $Q$ and it is again brought into contact with $S_2$ and removed. This procedure is repeated $n$ times.
(a) Find the electrostatic energy of $S_2$ after $n$ such contacts with $S_1$.
(b) What is the limiting value of this energy as $n 	o \infty$?

Sigma Percentile · Accepted Answer

The problem of repeatedly charging a sphere and bringing it into contact with another is a beautiful exploration of charge sharing, geometric progressions, and limits. Let's dive into the physics and math behind this fascinating process.

### Analyzing the Setup

Imagine we have two conducting spheres. Sphere $S_1$ has a radius $r$ and is attached to an insulating handle. It carries an initial charge $Q$. Sphere $S_2$ has a larger radius $R$ and sits on an insulating stand, initially uncharged.

We are going to repeatedly touch $S_1$ to $S_2$, recharging $S_1$ to $Q$ every single time. Before we start touching them, let's recall the golden rule of charge sharing. When two conductors touch, they reach a common potential. This means they share the total charge in the ratio of their capacitances.

For isolated spheres, capacitance is directly proportional to the radius ($C = 4\pi\varepsilon_0 R$). So, the charges will always divide in the ratio of $r$ to $R$.

### The First Contact

Let's make the first contact. $S_1$ brings a charge $Q$, and $S_2$ has zero. The total charge is $Q$. $S_2$ will take a fraction of this total charge based on its radius.

So, the charge on $S_2$ after the first contact, let's call it $q_1$, will be:

$$q_1 = Q \left( \frac{R}{R+r} \right)$$

### The Second Contact

Now, we pull $S_1$ away, recharge it back to $Q$, and touch it to $S_2$ again. This time, the total charge is $Q$ from $S_1$ plus $q_1$ already on $S_2$.

Again, $S_2$ takes its share, which is $\frac{R}{R+r}$ of the new total. If we substitute $q_1$, we get a beautiful expression with two terms:

$$q_2 = (Q + q_1) \left( \frac{R}{R+r} \right) = Q \left[ \frac{R}{R+r} + \left( \frac{R}{R+r} \right)^2 \right]$$

### The Geometric Progression

Do you see the pattern emerging? Every time we repeat this, we add another higher power term to the series. So, after $n$ such contacts, the charge $q_n$ on $S_2$ will be a sum of $n$ terms.

This is a classic Geometric Progression (GP), where the first term and the common ratio are both $\frac{R}{R+r}$:

$$q_n = Q \left[ \frac{R}{R+r} + \left( \frac{R}{R+r} \right)^2 + \dots + \left( \frac{R}{R+r} \right)^n \right]$$

Let's sum this GP. Using the standard formula for the sum of $n$ terms, $S_n = \frac{a(1 - x^n)}{1 - x}$, we substitute our first term and common ratio.

Notice how the denominator simplifies beautifully. $1 - \frac{R}{R+r}$ is just $\frac{r}{R+r}$. The $R+r$ terms cancel out, leaving us with a neat, compact formula for $q_n$:

$$q_n = Q \frac{R}{r} \left[ 1 - \left( \frac{R}{R+r} \right)^n \right]$$

### Final Calculation: Electrostatic Energy

Now for part (a) of the question. We need the electrostatic energy of $S_2$. The energy of a charged sphere is $\frac{q^2}{2C}$.

Since $C = 4\pi\varepsilon_0 R$, the energy is $\frac{q^2}{8\pi\varepsilon_0 R}$. We just plug in our expression for $q_n$, and we have our answer for the energy after $n$ contacts:

$$U_n = \frac{q_n^2}{8\pi\varepsilon_0 R} = \frac{Q^2 R}{8\pi\varepsilon_0 r^2} \left[ 1 - \left( \frac{R}{R+r} \right)^n \right]^2$$

### The Limiting Case

Finally, part (b) asks for the limiting value as $n$ approaches infinity. Look at the term $\frac{R}{R+r}$. Since $r$ is positive, this fraction is strictly less than $1$.

When you raise a fraction less than $1$ to an infinite power, it vanishes to zero! So, the maximum charge $S_2$ can hold is $Q \frac{R}{r}$, giving us the maximum limiting energy:

$$U_\infty = \frac{Q^2 R}{8\pi\varepsilon_0 r^2}$$

This is a brilliant problem that combines electrostatics with geometric progressions. Think about this: what if $S_1$ wasn't manually recharged, but instead kept connected to a constant voltage battery? How would the charge on $S_2$ grow then? Also, every time they touch, some energy is lost as heat. Calculating that total heat loss is a fantastic challenge for you to try next!

Visualized Solution

$Initial Setup$

$Charge Sharing Principle$

$First Contact$

$Second Contact$

$Pattern for n Contacts$

$Summing the GP$

$Electrostatic Energy after n Contacts$

$Limiting Case as n \to \infty$

$The Way Forward$

The Sigma Insight: Conductors

Analyzing the Setup

The First Contact

The Second Contact

The Geometric Progression

Final Calculation

The Limiting Case

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Initial Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Charge Sharing Principle

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } First Contact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Second Contact

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Pattern for n Contacts

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Summing the GP

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electrostatic Energy after n Contacts

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Limiting Case as n→∞

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward

The Sigma Insight: Conductors

Analyzing the Setup

The First Contact

The Second Contact

The Geometric Progression

Final Calculation

The Limiting Case

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Shown in the figure is a shell made of a conductor. It has inner radius a and outer radius b and carries charge Q. At its centre is a dipole p as shown. In this case,

$Initial Setup$

$Charge Sharing Principle$

$First Contact$

$Second Contact$

$Pattern for n Contacts$

$Summing the GP$

$Electrostatic Energy after n Contacts$

$Limiting Case as n \to \infty$

$The Way Forward$

Shown in the figure is a shell made of a conductor. It has inner radius $a$ and outer radius $b$ and carries charge $Q$ . At its centre is a dipole $p$ as shown. In this case,