Identical point charges of +2.2 µC are fixed to three of the four corners of a square. What is the magnitude of the negative point charge that must be fixed to the fourth corner, so that the charge at the diagonally opposite corner experiences a net force of zero?

1 Answer

  • 3 weeks ago

    Let square arm is of L meter and the value of the charge at fourth corner is -q µC,

    =>F by -qµC on the diagonally opposite charge = [k2.2q x 10^-12]/[√{2L^2}]^2 = [k2.2q x 10^-12]/2L^2 (attractive)

    Thus two component (equal in amount being a square) of F (attractive) in the direction of the other two corners shall be equal to the force applied (repulsive) by the two equal charges od 2.2µC,

    =>Fcos45* = Fsin45* = [k x 2.2 x 2.2 x 10^-12]/L^2

    =>[k2.2q x 10^-12]/2L^2 x 0.71 = [k x 2.2 x 2.2 x 10^-12]/L^2

    =>[q]/(2x 0.71) = [2.2]

    =>q = -3.124µC

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