What is the net electric force acting on Q1?

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  • 1 month ago

    The force on Q1 from Q2 is k.Q1.Q2/d²

    = 9*10^9 * 200*10^-6 * 100*10^-6 / 4.0² = 11.25N

    This is positive indicating repulsion. Q1 is pushed away from Q2.   So the force on Q1 from Q2 is 11.25N to the left.

    _____________________

    The force on Q1 from Q3 is k.Q1.Q3/d²= 9*10^9 * 200*10^-6 * (-50)*10^-6 / 6.0² = -2.5N

    This is negative indicating attraction.  Q1 is pulled towards Q3. So the force on Q1 from Q3 is 2.5N to the right.

    _____________________

    Net force = 11.25 - 2.5 = 8.75N to the left.

    If you are using (+x = right) and (-x = left) then the forces are -11.25N and 2.5N.  The net force is -11.25 + 2.5 = -8.75N

  • 1 month ago

    force due to Q2:

    F2 = (9e9)(200e-6)(100e-6) / (4)² = 180000e-3/16 = 180/16 = 11.25 N, to the left

    force due to Q3:F3 = (9e9)(200e-6)(50e-6) / (6)² = 90000e-3/36 = 90/16 = 5.63 N, to the right

    net is 5.63 N to the left

    Coulomb's law, force of attraction/repulsion

    F = kQ₁Q₂/r²

       Q₁ and Q₂ are the charges in coulombs

       F is force in newtons

       r is separation in meters

       k = 8.99e9 Nm²/C²

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