Anonymous asked in Science & MathematicsPhysics · 2 months ago

Electric Charge Physics?

A charge q1 of -5.00 × 10‐⁹ C and a charge q2 of -2.00 × 10‐⁹ C are separated by a distance of 40.0 cm. Find the equilibrium position for a third charge of +15.0 × 10‐⁹ C.

P.S. 10-⁹ is 10^-9

1 Answer

  • NCS
    Lv 7
    2 months ago
    Favorite Answer

    Since both charges have the same sign, the equilibrium point must be between them. Measured from q1, we need

    k*5.00e-9C / d² = k*2.00e-9C / (0.400m - d)²

    5.00*(0.400m - d)² = 2.00*d²

    This quadratic has a root at

    d = 0.245 m = 24.5 cm ◄ our solution

    and d = 1.09 m ← outside the permitted range

    Measured from q2, that would be 0.155 m = 15.5 cm.

    Note that the root d = 1.09 m is another point measured from q1 (and on the other side of q2) where the magnitudes of the fields are the same. But at that point, the fields both point toward both charges and so it is not an equilibrim point.

    Note also that ANY charge can be placed at the equilibrium point.

    Hope this helps!

    • Daniel2 months agoReport


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