Identical isolated conducting spheres 1 and 2, fixed in place, have equal charge

Question

Identical isolated conducting spheres 1 and 2, fixed in place, have equal charges +q and are separated by a distance that is large compared with their diameters (Figure a). The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F. Suppose now that a third identical sphere 3, having an insulating handle and initially having charge -q, is touched first to sphere 1 (Figure b), then to sphere 2 (Figure c), and finally removed (Figure d). The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is now f. What is the ratio f/F? A. 0 B. 1/2 C. 3/8 D. 3/4 E. 1

Explanation / Answer

Ans).

In the first figure (a),The force between the two sphere is by coulomb's law is

F =k .q1.q2/r2 =

since both are equal charges F =k. q2/r2   ---(1)

And in the 2nd figure(b), The positive charged sphere1 touched the negative charge sphere3 then the bodies become neutral

Hence force between them is Force =k.(+q+(-q)) q/(r2) =0 -----(2)

Now in the 3rd fugure (c), the positive charged sphere2 is touched with the neutral charged sphere3. Then here the charge redistribute and become half. i.e charge of 2nd sphere becomes 1/2. Because charge from 2nd sphere redistributed to 3rd sphere.

So f = k.q.(q/2)/r2  = k.(q2/2)/r2  -----(3)

Then the ration between f/F = (1)/(3)

f/F = (k.q2/r2)/(k.(q2/2)/r2)

f/F = q2/(q2/2)

f/F = 1/2

so finally the ratio of f/F is 1/2 Which means option B is correct.

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