Explore BrainMass

Explore BrainMass

    Electrostatic repulsion

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    A certain charge Q is divided into two parts q and Q-q. How the charges Q and q must be related so that when q and (Q-q) placed at a certain distance apart experience maximum electrostatic repulsion.

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf
    https://brainmass.com/physics/charge/electrostatic-repulsion-maximums-4061

    SOLUTION This solution is FREE courtesy of BrainMass!

    Ans. The electrostatic force of repulsion between the charges q and (Q-q) at separation r is given by,
    F = q(Q-q)/4peo r^2
    = (qq-q^2)/4peo r^2

    If F is maximum, then
    dF/dq = 0
    That is
    (Q-2q)/4peor^2 = 0

    As 1/4peor^2 is constant,
    Q-2q = 0
    Or Q = 2q

    Please see the word attachment for the well formatted answer.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:43 pm ad1c9bdddf>
    https://brainmass.com/physics/charge/electrostatic-repulsion-maximums-4061

    Attachments

    ADVERTISEMENT