Anonymous
Anonymous asked in Science & MathematicsPhysics · 3 months ago

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.119 N...?

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.119 N when their center-to-center separation is 71.9 cm.

The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0325 N. Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge on one of them and (b) the positive charge on the other? (Assume the negative charge has smaller magnitude.)

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  • NCS
    Lv 7
    3 months ago
    Favorite Answer

    electrostatic force F = kQq / d²

    where k = 8.99e9 N·m²/C²

    first condition:

    -0.119 N = 8.99e9N·m²/C² * Q * q / (0.719m)²

    Q*q = -6.84e-12 C²

    second condition:

    0.0325 N = 8.99e9N·m²/C² * Q'² / (0.719m)²

    Q'² = 1.87e-12 C²

    Q' = 1.367e-6 C

    and so the total charge present is Q_tot = 2.734e-6 C = Q + q

    and so

    q = 2.734e-6C - Q

    return to the first condition

    Q*(2.734e-6C - Q) = -6.84e-12 C²

    which has roots at

    Q = -1.58e-6 C = -1.58 µC ◄ initial negative

    and Q = 4.32e-6 C = 4.32 µC ◄ initial positive

    If you find this helpful, please award Best Answer!

  • JOHN
    Lv 7
    3 months ago

    Answer

    EDIT: Line 3 - the value for Q1Q2 should have a - sign for the reason given by NCS in Comments.

    Line 6 the first - sign should be changed to a + and changes mutatis mutandi made in the calculation. The answers will come out the same as NCS's.

    Attachment image
    • JOHN
      Lv 7
      3 months agoReport

      Yes, thanks NCS. I missed that. Too late, though, for an edit. BA should go to you.

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