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anonymous

  • one year ago

Two electrically-charged spheres are suspended from insulated threads a certain distance from each other. There is a certain amount of electrostatic force between them. Describe specifically (not just increase or decrease) what happens to this force in each of the scenarios below The charge on one sphere is reduced by half The charge on both spheres is doubled The distance between the spheres is increased by a factor of three The distance between the sphere is decreased to one-fourth The charge of each sphere is doubled and the distance between them is doubled

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  1. Abhisar
    • one year ago
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    brb

  2. Abhisar
    • one year ago
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    There?

  3. shamim
    • one year ago
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    Force between two electrically charged objects is F=k(q1*q2/r^2)

  4. shamim
    • one year ago
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    Frm my equation If charge of any 1 sphere is reduced to half then force between them will b half of the previous force

  5. Michele_Laino
    • one year ago
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    let's suppose that Q_1 is the charge on sphere #1, and Q_2 is the charge on sphere #2, then the electrostatic force has the subsequent magnitude: \[F = k\frac{{{Q_1}{Q_2}}}{{{d^2}}}\] where d is the distance between the spheres

  6. Michele_Laino
    • one year ago
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    first scenarios: the new charge on sphere #1 is Q_1/2, so we have: \[{F_1} = k\frac{{\left( {{Q_1}/2} \right){Q_2}}}{{{d^2}}} = ...\]

  7. Michele_Laino
    • one year ago
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    scenario*

  8. Michele_Laino
    • one year ago
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    second scenario: the new charges are: 2Q_1 and 2Q_2 on the spheres #1 and #2 respectively. The new force is: \[{F_2} = k\frac{{{{\left( {2Q} \right)}_1}\left( {2{Q_2}} \right)}}{{{d^2}}} = ...\]

  9. Michele_Laino
    • one year ago
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    third scenario: the new distance is 3d, so the new force is: \[{F_3} = k\frac{{{Q_1}{Q_2}}}{{{{\left( {3d} \right)}^2}}} = ...\]

  10. Michele_Laino
    • one year ago
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    fourth scenario: the new distance is d/4, so the electrical force is: \[{F_4} = k\frac{{{Q_1}{Q_2}}}{{{{\left( {d/4} \right)}^2}}} = ...\]

  11. Michele_Laino
    • one year ago
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    fifth scenario: the new charges are 2Q_1 and 2Q_2 on the spheres #1 and #2 respectively, furthermore the new distance is 2d, so the electrostatic force is: \[{F_5} = k\frac{{{{\left( {2Q} \right)}_1}\left( {2{Q_2}} \right)}}{{{{\left( {2d} \right)}^2}}} = ...\]

  12. Michele_Laino
    • one year ago
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    hint: in each of five scenarios, you have to express the new electrostatic force, namely F_1, F_2, F_3, F_4, and F_5 in terms of the force F

  13. anonymous
    • one year ago
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    ie. F_1 is one-half the magnitude initial electrostatic force, F_2 is four times the magnitude initial electrostatic force, F_3 is one-ninth the magnitude initial electrostatic force, F_4 is sixteen times the magnitude initial electrostatic force and F_5 is equal in magnitude to the initial electrostatic force.

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