Dear Sir,

Two metal spheres A and B of radius r and 2r whose centres are separated by a distance of 6r are given charge Q are at potential V1 and V2. Find the ratio of V1/V2. These spheres are connected to each other with the help of a connecting wire keeping the separation unchanged, with its the amount of charge that will flow through the wire?

To tackle this problem, we need to analyze the electric potentials of the two spheres and how they interact when connected by a wire. Let's break it down step by step.

Understanding Electric Potential

The electric potential \( V \) of a charged sphere is given by the formula:

\( V = \frac{kQ}{r} \)

where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the radius of the sphere. For our two spheres, we have:

• Sphere A with radius \( r \) and charge \( Q \)
• Sphere B with radius \( 2r \) and charge \( Q \)

Calculating the Potentials

Now, let's calculate the potentials \( V_1 \) and \( V_2 \) for spheres A and B respectively:

• For sphere A:

\( V_1 = \frac{kQ}{r} \)

• For sphere B:

\( V_2 = \frac{kQ}{2r} \)

Finding the Ratio of Potentials

To find the ratio \( \frac{V_1}{V_2} \), we can substitute the expressions we derived:

\( \frac{V_1}{V_2} = \frac{\frac{kQ}{r}}{\frac{kQ}{2r}} \)

By simplifying this, we get:

\( \frac{V_1}{V_2} = \frac{2r}{r} = 2 \)

Result of the Ratio

Thus, the ratio of the potentials is:

2

Charge Flow When Connected

Now, let’s consider what happens when the two spheres are connected by a wire. When connected, charge will flow until both spheres reach the same potential.

Equalizing the Potentials

Let \( Q_A \) and \( Q_B \) be the charges on spheres A and B after they are connected. The total charge remains the same:

\( Q_A + Q_B = 2Q \)

The potentials of the spheres after connection can be expressed as:

• For sphere A: \( V_A = \frac{kQ_A}{r} \)
• For sphere B: \( V_B = \frac{kQ_B}{2r} \)

Setting these equal gives us:

\( \frac{kQ_A}{r} = \frac{kQ_B}{2r} \)

From this, we can derive:

\( 2Q_A = Q_B \)

Substituting Back

Now substituting \( Q_B \) in terms of \( Q_A \) into the total charge equation:

\( Q_A + 2Q_A = 2Q \)

Thus:

\( 3Q_A = 2Q \)

Solving for \( Q_A \):

\( Q_A = \frac{2Q}{3} \)

And substituting back to find \( Q_B \):

\( Q_B = 2Q_A = \frac{4Q}{3} \)

Charge Flow Calculation

The initial charge on sphere A was \( Q \), and after connection, it becomes \( \frac{2Q}{3} \). Therefore, the charge that flows from sphere A to the wire is:

\( Q - Q_A = Q - \frac{2Q}{3} = \frac{Q}{3} \)

In summary, the charge that flows through the wire when the spheres are connected is:

\( \frac{Q}{3} \)

This analysis shows how electric potential and charge distribution work in conductive systems, providing a clear understanding of the behavior of charged spheres when connected. If you have any further questions or need clarification on any part, feel free to ask!