Question

Two identical conducting spheres were situated so that they were insulated from their surroundings, and their centers were separated by a distance, d.
One sphere had a charge of $+8$ Coulombs; the other had a charge of $-4$ Coulombs.
The spheres were then brought together to touch each other. The spheres were separated and again insulated from their surroundings, and their centers were separated by a distance d.
How does the magnitude of the electrostatic force between the spheres in their final situation compare to the magnitude of electrostatic force between them in their original situation?

• A. The magnitude of the final force is the same as the magnitude of the original force
• B. The magnitude of the final force is half as much as the magnitude of the original force
• C. The magnitude of the final force is twice as much as the magnitude of the original force
• D. The magnitude of the final force is one-fourth as much as the magnitude of the original force
• E. The magnitude of the final force is one-eighth as much as the magnitude of the original force

Answer 1

The correct option is E The magnitude of the final force is one-eighth as much as the magnitude of the original force

Before touch , the electrostatic force between two given identical spheres ,

$F_{before}=k8\times 4/d^2$ ...............................eq1

net charge of spheres $=+8-4=+4C$

now as the spheres are identical therefore their capacitances will be same , we know charges on spheres after touch are given by ,

$q_1^{\prime }/q_2^{\prime }=C_1/C_2=1$ (because $C_1=C_2$)

or $q_1=q_2$

it means net charge will be divided equally on both he spheres i.e. $q_1=q_2=4/2=2C$

therefore after touch , the electrostatic force between two spheres ,

$F_{afer}=k2\times 2/d^2$ ...............................eq2

dividing eq2 by eq1 ,

$F_{after}/F_{before}=4/32=1/8$

or $F_{after}=1/8F_{before}$