Two similar sphere having charge -2C and +4C respectively are separated by distance "r" exert force (F=16N) third similar neutral sphere is first touched with sphere B and then sphere A. then force between spheres decreases by:

Answer 1

When the third neutral sphere is first touched with sphere B (+4C), the charges are shared equally between them, so each sphere has a charge of +2C. Then, when the neutral sphere (now charged with +2C) is touched with sphere A (-2C), the charges are again shared equally between them, and both spheres become neutral (0C).

Now, the force between the two spheres is given by Coulomb's Law:

F = k * (q1 * q2) / r^2

Where F is the force, k is Coulomb's constant, q1 and q2 are the charges on the two spheres, and r is the distance between them.

Initially, the force between the two spheres was 16 N, so:

16 = k * (-2C * +4C) / r^2

After touching the neutral sphere, both sphere A and sphere B become neutral, so the force between them will be 0, as they have no charge.

The decrease in force can be calculated as:

Decrease in force = Initial force - Final force
Decrease = 16 - 0
Decrease = 16 N