Two spherical conductors B and C having equal radii and carrying equal charges in them repel each ot

Q: Why do the spheres share the charge exactly equally?

Because the spheres are identical, they have the same capacitance. When they touch, charge flows until their electrical potentials are equal. For identical spheres, equal potential means equal charge.

Q: What would happen if the spheres had different sizes?

If the spheres had different radii, they would not share the charge equally. Instead, the charge would distribute itself in direct proportion to their radii (or capacitance) to equalize the potential.

Q: Does the distance between sphere B and C change during the process?

No, the problem states they are 'kept apart at some distance', implying their positions are fixed. Only sphere A is moved around to transfer the charge.

Q: Why is the final force still repulsive?

Because we started with equal charges (let's assume positive), and we only shared this positive charge around. Since both B and C end up with positive charges ($\frac{q}{2}$ and $\frac{3q}{4}$), they will still repel each other.

Q: Where does the 'missing' charge go at the end?

It's not missing! Sphere A is removed from the system at the end, and it takes its final charge of $\frac{3q}{4}$ with it. If you add up the final charges of A, B, and C, it will equal the initial total charge of $2q$.

Question

Two spherical conductors $B$ and $C$ having equal radii and carrying equal charges in them repel each other with a force $F$ when kept apart at some distance. A third spherical conductor having same radius as that of $B$ but uncharged, is brought in contact with $B$, then brought in contact with $C$ and finally removed away from both. The new force of repulsion between $B$ and $C$ is

Options:

(A) $\frac{F}{4}$

(B) $\frac{3F}{4}$

(C) $\frac{F}{8}$

(D) $\frac{3F}{8}$

Sigma Percentile · Accepted Answer

Have you ever wondered how nature perfectly balances its books? In the world of electrostatics, charge is a fundamental currency. It can neither be created nor destroyed, only transferred. This problem is a beautiful, classic demonstration of that exact principle—the conservation of charge—combined with the elegant symmetry of identical conductors.

Imagine you are standing in a lab, looking at two identical spherical conductors, $B$ and $C$. They are suspended at a fixed distance $r$ from each other.

### The Initial State

Let's analyze what we have right at the beginning. Both spheres carry an identical charge, which we will call $q$. Because they are identically charged, they don't want to be near each other. They push each other away with a repulsive electrostatic force, $F$.

According to **Coulomb's Law**, the magnitude of this force is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

$$F = \frac{1}{4\pi\varepsilon_0} \frac{q \cdot q}{r^2} = \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2}$$

This equation is our baseline. It is the reference point we will use to compare our final result. Keep this safely tucked away in your mind.

### The First Encounter

Now, the plot thickens. We introduce a third sphere, $A$. This sphere is the exact identical twin of $B$ and $C$ in terms of its size and shape. However, it is completely uncharged. Its initial charge is a perfect zero ($q_A = 0$).

We take this neutral sphere $A$ and bring it into direct physical contact with sphere $B$. What happens at the microscopic level?

Because the spheres are identical conductors, they have the exact same capacity to hold charge. Nature loves symmetry and equilibrium. The moment they touch, the charges redistribute themselves until both spheres reach the exact same electrical potential. For identical spheres, this simply means they split the total available charge equally down the middle.

The total charge before they touch is the charge of $A$ plus the charge of $B$: $0 + q = q$.

When they separate, they share this total charge equally:

$$q_A' = q_B' = \frac{0 + q}{2} = \frac{q}{2}$$

Sphere $B$ has now lost half of its charge to $A$. Both walk away from this encounter carrying a charge of $\frac{q}{2}$.

### The Second Encounter

Sphere $A$ is no longer neutral. It is now carrying a charge of $\frac{q}{2}$. But its journey isn't over. We now take sphere $A$ and bring it into contact with sphere $C$.

Sphere $C$ has been patiently waiting, still holding onto its original full charge, $q$.

Once again, the rule of identical conductors applies. They will pool their resources and split the total charge equally. Let's find out what that total pool is.

The total charge is now the new charge of $A$ plus the charge of $C$:

$$	ext{Total Charge} = \frac{q}{2} + q = \frac{3q}{2}$$

When they separate, this total charge is divided equally by two:

$$q_A'' = q_C' = \frac{\frac{3q}{2}}{2} = \frac{3q}{4}$$

Sphere $C$ has now shared some of its charge with $A$. Sphere $C$ is left with a charge of $\frac{3q}{4}$, and sphere $A$ walks away with the same amount.

### The Final Calculation

Finally, sphere $A$ is removed from the scene entirely. It takes its $\frac{3q}{4}$ charge with it, leaving $B$ and $C$ to interact once again.

Let's take stock of our final setup. Sphere $B$ now has a charge $q_B' = \frac{q}{2}$. Sphere $C$ now has a charge $q_C' = \frac{3q}{4}$. The distance between them, $r$, has not changed.

We need to find the new force of repulsion, $F'$, between them. We return to our trusty tool, **Coulomb's Law**:

$$F' = \frac{1}{4\pi\varepsilon_0} \frac{q_B' \cdot q_C'}{r^2}$$

Let's carefully substitute our new charge values into the equation:

$$F' = \frac{1}{4\pi\varepsilon_0} \frac{\left(\frac{q}{2}ight) \left(\frac{3q}{4}ight)}{r^2}$$

Now, let's do a bit of algebraic housekeeping. We can pull the numerical fractions out to the front:

$$F' = \frac{3}{8} \left( \frac{1}{4\pi\varepsilon_0} \frac{q^2}{r^2} ight)$$

Look closely at the term inside the parentheses. Does it look familiar? It should! It is the exact expression for our initial force, $F$, that we established right at the beginning.

Therefore, we can elegantly substitute $F$ back into our equation:

$$F' = \frac{3F}{8}$$

And there we have it! The new force of repulsion is exactly $\frac{3}{8}$ of the original force.

This problem is a fantastic reminder of how methodical, step-by-step analysis can unravel seemingly complex interactions. By simply following the charge and applying the principle of equal sharing for identical conductors, we arrived at a clean, beautiful mathematical conclusion.

Select Answer:

Visualized Solution

$F_{ini t ia l}$

$q_{A}$

$A \to B$

$q_{A}^{'}, q_{B}^{'}$

$A \to C$

$q_{C}^{'}$

$F_{f ina l}$

$F^{'}$

\text{Result}

The Sigma Insight: Coulomb's Law

The Initial State

The First Encounter

The Second Encounter

The Final Calculation

Similar Questions

Three charges $+ Q, q, + Q$ are placed respectively at distance $0, d /2$ and $d$ from the origin on the X-axis. If the net force experienced by $+ Q$ placed at $x = 0$ is zero, then value of $q$ is

A charge $q$ is placed at the centre of the line joining two equal charges $Q$ . The system of the three charges will be in equilibrium if $q$ is equal to

A certain charge $Q$ is divided into two parts $q$ and $(Q - q)$ . How should the charges $Q$ and $q$ be divided, so that $q$ and $(Q - q)$ placed at a certain distance apart experience maximum electrostatic repulsion ?

Three charges $- q_{1}, + q_{2}$ and $- q_{3}$ are placed as shown in the figure. The x-component of the force on $- q_{1}$ is proportional to

If a charge $q$ is placed at the centre of the line joining two equal charges $Q$ such that the system is in equilibrium, then the value of $q$ is

A charge $Q$ is placed at each of the opposite corners of a square. A charge $q$ is placed at each of the other two corners. If the net electrical force on $Q$ is zero, then $Q / q$ equals

Five point charges, each of value $+ q$ coulomb, are placed on five vertices of a regular hexagon of side $L$ metre. The magnitude of the force on the point charge of value $- q$ coulomb placed at the centre of the hexagon is ………newton.

Select Answer:

Visualized Solution

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qA′​,qB′​

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

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

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

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