Question

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters. The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 0.42 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere , then to sphere 2, and finally removed. What is the magnitude of the electrostatic force F' that now acts on sphere 2?

Answer 1

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = $\frac{kQ_{1}Q_{2} }{r^{2} }$

F= $\frac{K(Q/2)(3Q/4)}{r^{2} }$

how $\frac{KQ^{2} }{r^{2} }$ = 0.42

Then

F = $\frac{0.42*3}{8}$

F = 0.1575 N