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20 March, 23:56

A 4.89 μC test charge is placed 4.10 cm away from a large, flat, uniformly charged nonconducting surface. The force on the charge is 321 N. The charge is then moved 2.00 cm farther away from the surface. What is the force on the test charge now?

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  1. 21 March, 02:49
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    The new force on the test charge is 145.02N

    Explanation:

    Force on a unit positive charge can be calculated using coulomb's law.

    F = Kq²/r²

    Where

    F is the force on the charge = 321 N

    K is a constant = 8.99 X10⁹ Nm²/C²

    q is a test charge = 4.89 μC = 4.89 X10⁻⁶ C

    r is the distance between the charge and the surface = 4.1cm

    F ∝1/r²

    Force on the charge is inversely proportional the square of distance between the charge and the surface.

    Fr² = constant

    F₁r₁² = F₂r₂²

    F₂ = F₁r₁²/r₂²

    If the charge is then moved 2.00 cm farther away from the surface;

    The new distance becomes 4.10 cm + 2.00 cm = 6.1 cm

    Therefore, r₂ = 6.1 cm, r₁ = 4.1 cm, F₁ = 321 N

    F₂ = (321 X4.1²) / (6.1²)

    F₂ = 145.02N

    Therefore, The new force on the test charge is 145.02N
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