Problem 164 - A Complete Resource Book in Physics for JEE Main

Question

Two spherical conductors \(A\) and \(B\) of radii \(1 \mathrm{~mm}\) and \(\mathrm{mm}\) are separated by a distance of \(5 \mathrm{~cm}\) and are uniformly charged. If the spheres are connected by conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres \(A\) and \(B\) is (A) \(4: 1\) (B) \(1: 2\) (C) \(2: 1\) (D) \(1: 4\)

Step-by-Step Solution

Answer

The ratio of the magnitude of electric fields at the surfaces of spheres A and B in equilibrium condition, when connected by a conducting wire is 2:1. The correct answer is (C) \(2: 1\).

Step 1: Formulate electric field expressions for sphere A and B

The electric field at the surface of sphere A is given by \(E_A = \frac{KQ_A}{r_A^2}\) and sphere B is given by \(E_B = \frac{KQ_B}{r_B^2}\)

Step 2: Write down the condition for equilibrium

Sphere A and B are in equilibrium when they are connected by a conducting wire, which means the potential across A and B will be the same i.e \( V_A = V_B\). The potential at the surface is given by \(V = \frac{KQ}{r}\) , substituting this in the equilibrium condition we get \(\frac{KQ_A}{r_A} = \frac{KQ_B}{r_B}\)

Step 3: Solve for the ratio of charges \(Q_A/Q_B\)

From step 2 we can solve for \(Q_A/Q_B = r_A/r_B = 1/2\)

Step 4: Calculate the ratio of electric fields \(E_A/E_B\)

Substitute \(Q_A/Q_B\) from step 3 into the expression for the ratio of electric fields we derived in step 1: \(E_A/E_B = Q_A/Q_B \times (r_B/r_A)^2\) , which simplifies to \(E_A/E_B = (1/2) \times (2/1)^2 = 2/1\)

Step 5: Result

The ratio of the magnitude of electric fields at the surfaces of spheres A and B in equilibrium condition, when connected by a conducting wire is 2:1 . The correct answer is (C) \(2: 1\).

Key Concepts

Conductors

Conductors are materials that allow electric charges to flow freely through them. This is because they have free electrons or other charge carriers that are not tightly bound to atoms.
In this context, when two spherical conductors are connected by a wire, they allow charges to redistribute until they reach an equilibrium condition.
This means that current flows between the spheres until the potential difference between them is zero.

• Conductors adjust until the electric field inside them is zero.
• The surface charge can move to equalize potential across both bodies.
• This property is why charges redistribute evenly when joined by a wire.

Understanding conductors helps explain the behavior of electric fields in such systems.

Equilibrium Condition

In the study of electric fields and conductors, equilibrium is reached when there is no net movement of charge within the conductor. This happens because the electric potential across interconnected conductors becomes equal.
For the two spheres in the scenario, they are said to be in equilibrium when they have the same electric potential.
Mathematically, the potential at the surface of a sphere is given by the formula \[ V = \frac{KQ}{r} \] where \( K \) is a constant, \( Q \) is the charge, and \( r \) is the radius.

• Equilibrium ensures both spheres have the same potential.
• Charge distributions adjust until this condition is satisfied.

This property leads to a specific ratio of the charges on the spheres, and consequently their electric fields as well.

Spherical Charge Distribution

Spherical charge distribution refers to the uniform arrangement of charge on a spherical object. For the two spheres in this exercise, the charge distribution affects the electric field on their surfaces. The electric field, denoted by \( E \), on a sphere's surface is given by \[ E = \frac{KQ}{r^2} \]where \( Q \) is the charge and \( r \) is the sphere's radius.
As the radius changes, the distribution modifies how the field behaves.

• The electric field magnitude is directly proportional to the charge \( Q \).
• It is inversely proportional to the square of the radius \( r \).

The ratio of fields for two connected spheres depend on their charge distributions and radii, as demonstrated by the problem's solution.