Question

Two uniformly charged spherical conductors, A and B of radii 5 mm and 10 mm are separated by a distance of 2 cm. If the spheres are connected by a conducting wire, then in equilibrium condition, the ratio of the magnitude of the electric fields at surface of the spheres A and B will be

• A. 1:2
• B. 2:1
• C. 1:1
• D. 1:4

Answer 1

The Correct Option is B

To solve this problem, we need to understand the concept of electric fields around spherical conductors when they are in equilibrium and connected by a conducting wire.

When two conductors are connected by a conducting wire, they come to the same potential in equilibrium. The potential \( V \) on the surface of a spherical conductor is given by the formula:

V = \frac{kQ}{r}

where \( k \) is Coulomb's constant, \( Q \) is the charge on the sphere, and \( r \) is the radius of the sphere.

For equilibrium conditions, we have:

V_A = V_B

Substituting the formula for potential, we get:

\frac{kQ_A}{r_A} = \frac{kQ_B}{r_B}

Since the spheres are in equilibrium, their potentials are equal and the constant \( k \) cancels out. This results in:

\frac{Q_A}{r_A} = \frac{Q_B}{r_B}

The electric field \( E \) at the surface of any spherical conductor is given by:

E = \frac{Q}{4\pi \varepsilon_0 r^2}

At equilibrium, when calculating the ratio of the electric fields at the surface of spheres A and B, we have:

\frac{E_A}{E_B} = \frac{\frac{Q_A}{4\pi \varepsilon_0 r_A^2}}{\frac{Q_B}{4\pi \varepsilon_0 r_B^2}} = \frac{Q_A \cdot r_B^2}{Q_B \cdot r_A^2}

Substitute the relation from potential equality:

\frac{E_A}{E_B} = \frac{r_B \cdot r_B^2}{r_A \cdot r_A^2} = \frac{r_B}{r_A}

Given r_A = 5\, \text{mm} and r_B = 10\, \text{mm}, we substitute these values:

\frac{E_A}{E_B} = \frac{10}{5} = 2

Thus, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is 2:1.