Ask Question
23 December, 17:27

A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apart. For the force that each object exerts on the other to be a maximum, what should q be? a) q 20 c) q Q/2 d) q / 4

+1
Answers (1)
  1. 23 December, 20:13
    0
    q = Q/2

    option c is correct

    Explanation:

    given data

    charge = Q

    distance r = 1 m

    charge = q

    to find out

    what should q be

    solution

    we have given charge q is remove so

    q1 = Q-q

    q2 = q

    we know by coulombs law force

    force = kq1q2 / r² ... 1

    put here value and we know k is constant

    F = k (Q-q) q / 1

    now take derivative charge q and put = 0

    dF/dq = k d (Q-q) q / dq

    so k (Q-q) q = 0

    Q = 2q

    q = Q/2

    so option c is correct
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “A small object has charge Q. Charge q is removed from it and placed on a second small object The two objects are placed I m apart. For the ...” 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