Question

The radii of two metallic spheres A and B are r1 and r2 respectively (r1 > r2). They are connected by a thin wire and the system is given a certain charge. The charge will be greater
(a) on the surface of the sphere B.
(b) on the surface of the sphere A.
(c) equal on both.
(d) zero on both.

Answer 1

Explanation:

The radii of two metallic spheres A and B are r1 and r2 respectively (r1 > r2). They are connected by a thin wire and the system is given a certain charge. The charge will be greater

Answer 2

Option (b) is right.

The charge will be greater on the surface of sphere A.

• The two spheres A and B have radii R1 and R2, where R1 > R2
• Since both are connected by a thin wire, they attain the same potential.
• The potential of a sphere is given by :
•      V = $\frac{KQ}{R}$
• Now the system is provided with a certain charge, the above equation can be arranged as:
•       $\frac{KQ1}{R1}$ = $\frac{KQ2}{R2}$
• i.e      $\frac{Q1}{R1}$ = $\frac{Q2}{R2}$
• But, we know  R1 > R2
• ∴   Q1 > Q2
• i.e charge on sphere A > charge on sphere B
• So we can conclude that the charge on sphere A is greater