Question

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

Short answer

The ratio of the magnitude of the electric fields at the surfaces of spheres A and B is \(2:1\).

Step-by-step solution

Understanding the equilibrium condition

In equilibrium condition, the potential of conductors connected by a conductor remains the same. For spheres A and B, this means that the potential on A is equal to the potential on B. Therefore, we have: \(ER_{A} = ER_{B}\) or \(E_{A} = \frac{R_{B}}{R_{A}}E_{B}\) (1)

Calculate the ratio of the electric fields

The ratio of the magnitude of the electric fields at the surfaces of spheres A and B can be calculated by substituting the radius of sphere A \(R_{A}=1mm\) and B \(R_{B}=2mm\) into equation (1). The result is \(E_{A}/E_{B} = \frac{2mm}{1mm} = 2\). The correct choice is (C) \(2: 1\). So, the ratio of the electric fields at the surfaces of the two spheres (A and B) is \(2:1\). This suggests that the electric field at the surface of sphere A is twice as strong as that at the surface of sphere B. This is due to the smaller radius of sphere A, which results in a stronger electric field.

Key concepts

Charged Spherical Conductors

Charged spherical conductors are fascinating objects in electrostatics, primarily because they distribute charge uniformly on their surface. When spherical conductors are connected by a wire, they move towards an equilibrium state, meaning they adjust until their electric potentials are equal. This happens because charges flow freely until they no longer experience a potential difference, effectively "balancing out."

For our conductors A and B, it is crucial to understand that their final state depends on both the charge distribution and physical properties like radius. Charge tends to stay on the surface due to repulsion of like charges. Hence, the shape and size of the spheres affect how the electric fields behave on their surfaces. A smaller sphere will generally have a stronger electric field on its surface compared to a larger sphere with the same total charge.

Equilibrium Condition

The equilibrium condition for two connected conductors can be simplified to equal electric potentials for both conductors. It indicates that any potential difference initially present gets neutralized by the flow of charge. This is because the charge in a conductor will move until the electric potential throughout the conductor is uniform.

For spheres connected with a wire, as in our example, this equilibrium ensures that the charge redistributes until the potentials on both surfaces are the same. This leads to the relationship between the electric fields and radii of the spheres. Given the formula \( E_A = \frac{R_B}{R_A}E_B \), we see how the radius influences the field's strength, maintaining the same potential across both conductors.

Electric Potential

Electric potential is a measure of the electric potential energy per unit charge at a point in a field. It represents the work done to move a charge from a reference point to a specific point within the field, without any acceleration. For spherical conductors, the electric potential is directly related to both the charge and radius of the sphere.

When the potentials of spheres A and B are equal, the potential \( V \) can be derived as \( V_A = V_B = k \frac{Q_A}{R_A} = k \frac{Q_B}{R_B} \), where \( k \) is Coulomb's constant and \( Q \) is the charge. This shows that smaller radii, like that of sphere A, result in a higher electric potential if the net charge remains constant. This is why field strengths differ even when potentials equalize.

Electric Field Ratio

The electric field ratio between two connected charged spherical conductors can be surprising if one doesn't consider the impact of radius. The formula \( \frac{E_A}{E_B} = \frac{R_B}{R_A} \) highlights how the ratio of surface electric fields is inversely proportional to the ratio of their radii.

Since sphere A is physically smaller than sphere B, the electric field at its surface is stronger. Our calculated result, \( 2:1 \), is derived from substituting the radii: \( \frac{E_A}{E_B} = \frac{2mm}{1mm} = 2 \). Thus, sphere A's surface electric field is twice that of sphere B. This arises because the same potential across differing radii requires a stronger field where the radius is smaller, as potential depends linearly on charge and inversely on radius.