Animated Solution for Physics - Electrostatics: Two isolated conducting spheres $S_1$ and $S_2$ of radii $\frac{2}{3}R$ and $\frac{1}{3}R$, have charges $12\mu \text{C}$ and $-3\mu \text{C}$ respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done, the charges on $S_1$ and $S_2$ respectively, are

Achieving Electrostatic Equilibrium

Imagine you are observing two isolated worlds, each holding its own electrical secrets. In the vast expanse of space, we have two conducting spheres, $S_1$ and $S_2$. They are separated by a distance so large that they are completely oblivious to each other's electric fields. This is a classic and beautiful problem in electrostatics that tests our understanding of potential, capacitance, and the conservation of charge.

Let's break down the initial state of our system. The first sphere, $S_1$, is the larger of the two, boasting a radius of $R_1=\frac{2}{3}R$. It carries a substantial positive charge, $Q_1=12\mu \text{C}$. The second sphere, $S_2$, is smaller, with a radius of $R_2=\frac{1}{3}R$, and it carries a negative charge, $Q_2=-3\mu \text{C}$.

Because they are isolated and far apart, we can treat them as independent spherical capacitors. The capacitance of an isolated conducting sphere is directly proportional to its radius, given by the elegant formula $C=4\pi {\epsilon }_{0}R$.

A Bridge Between Worlds

Now, the magic happens. We introduce a conducting wire and connect the two spheres. Suddenly, these isolated worlds are linked. But what exactly happens when this connection is made?

Think of electrical potential as the "electrical pressure" or "water level" in a tank. Just as water flows from a higher level to a lower level through a connecting pipe, electric charge flows from a region of higher potential to a region of lower potential through the conducting wire.

The charge will continue to flow until the "electrical pressure" equalizes. In physics terms, the flow of charge ceases when both spheres reach the exact same electrostatic potential. We call this the Common Potential ($V_C$). Once this state is reached, the system is in electrostatic equilibrium.

Finding the Common Potential

To find this common potential, we rely on two fundamental principles: the conservation of charge and the definition of capacitance.

Even though charge moves from one sphere to the other, the total charge of the isolated system remains strictly conserved. No charge is created or destroyed; it is merely redistributed.

The common potential $V_C$ can be found by dividing the total charge of the system by the total equivalent capacitance. Since the spheres are connected by a wire, they are effectively in a parallel combination.

Let's calculate the total charge first. We must be careful with the signs!

Next, we determine the total capacitance. We substitute the formula for the capacitance of a sphere:

Now, we substitute the given radii into our capacitance equation:

Isn't it beautiful how the fractions perfectly add up to $1$? This gives us a clean expression for the common potential:

Where Did the Charges Go?

Now that we know the common potential, finding the final charge on each sphere is a breeze. The final charge is simply the sphere's individual capacitance multiplied by the common potential.

Let's calculate the final charge on the first sphere, $Q_1^{\prime }$:

Notice how the $4\pi {\epsilon }_{0}R$ terms gracefully cancel out! This leaves us with a simple arithmetic calculation:

Similarly, let's find the final charge on the second sphere, $Q_2^{\prime }$:

Again, the complex terms cancel out, revealing the final charge:

The Charge Sharing Ratio

While calculating the common potential is a robust method, there is a faster, more intuitive way to solve this.

Since both spheres reach the same potential ($V=\text{constant}$), and we know that $Q=CV$, it implies that the final charge is directly proportional to the capacitance. For spheres, capacitance is directly proportional to the radius.

We need to divide the total charge of $9\mu \text{C}$ into a $2:1$ ratio. - The first sphere gets $\frac{2}{3}$ of the total charge: $\frac{2}{3}\times 9\mu \text{C}=6\mu \text{C}$. - The second sphere gets $\frac{1}{3}$ of the total charge: $\frac{1}{3}\times 9\mu \text{C}=3\mu \text{C}$.

This confirms our previous calculation and provides a lightning-fast technique for competitive exams! The final charges are indeed $6\mu \text{C}$ and $3\mu \text{C}$, making option (d) the correct choice.