Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters. A third id

Question

Two identical conducting spheres, A and B, carry equal charge. They are separated by a distance much larger than their diameters. A third identical conducting sphere, C, is uncharged. Sphere C is first touched to A, then to B, and finally removed.
1. As a result, the electrostatic force between A and B, which was originally F, becomes ________:A. F/2B. F/4C. 3F/8D. F/16E. 0

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Jezebel 4 years 2021-08-27T10:53:34+00:00 1 Answers 41 views 0

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  1. Vodka
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    2021-08-27T10:54:37+00:00
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    Answer:

    C. \frac{3F}{8}

    Explanation:

    Let initial charges on both spheres be,q

    F=\frac{Kq^2}{d^2} \ \ \ \ \ \ \ \ \ \ \_i

    When the sphere C is touched by A, the final charges on both will be,\frac{q}{2}

    #Now, when C is touched by B, the final charges on both of them will be:

    q_c=q_d=\frac{q/2+q}{2}\\\\=\frac{3q}{4}\\

    Now the force between A and B is calculated as:

    F\prime=\frac{k\times\frac{q}{2}\times \frac{3q}{4}}{d^2}\\F\prime=\frac{3F}{8}

    Hence the electrostatic force becomes 3F/8

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