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 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 1, then to sphere 2 , and finally removed. The electrostatic force that now acts on sphere 2 has magnitude F .What is the ratio F /F?

Answer 1

0.375

Step-by-step explanation:

When the 3rd sphere touches the 1st one, the charge will then be distributed between both of them, then now the 1st sphere has only half of his original charge.

In this moment then

Sphere 1 has a charge = Q/2

Sphere 3 has a charge = Q/2

When the 3rd sphere touches the 2nd 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 2 has a charge = 3/4Q

Sphere 3 has a charge = 3/4Q

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

F = (kq1q2) / r²

F = (Q/2 * 3Q/4) / r²

F = (Q² * 3) / 8r²

From the question, F = 0.42 = kQ²/r²

Thus, we can say that

F = (0.42 * 3) / 8

F = 0.1575

Thus, the ratio between F/F =

0.1575 / 0.42

Ratio, r = 0.375