Ask Question
30 July, 18:39

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?

+2
Answers (1)
  1. 30 July, 20:18
    0
    0.375

    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
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers