Identical isolated conducting spheres 1 and 2 have equal charges and are separat

Question

Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with their diameters (see Figure (a)). The magnitude of the electrostatic force acting on sphere 2 due to sphere 1 is F = 5.6 N. Suppose now that a third identical sphere 3, having an insulating handle and initially neutral, is touched first to sphere 1 (see Figure (b)), then to sphere 2 (see Figure (c)), and finally removed (see Figure (d)). What is the magnitude of the electrostatic force F' that now acts on sphere 2?

Explanation / Answer

let us assume , initially sphere 1 and sphere 2 has charge = q

sphere 3 has no charge.

The magnitude of electrostatic force between sphere 1 and 2:

F = k q1*q2 / r2   ( r = distance betwwen charges 1 and2)

= k q*q / r2

F = k q2 / r2   -------------------------------1

Now when the sphere 3 is touched with sphere 1 .Because sphere are exactly same they will end up with same charge.

Now charge on sphere 1 and sphere 3 = q/2

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Now sphere 3 is touched with sphere 2.

Total charge on two spheres = q + q/2 = 3q/2

Now charge again will distribute evenly between two spheres .

Now charge on sphere 2 and sphere 3 = (3q/2) / 2

= 3q/4

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the magnitude of final force between 1 and 2

F' = k q1*q2 / r2

= k (q/2)* ( 3q/4) / r2

= k 3q2 / 8 r2   ( from eq 1, F = k q2 / r2)

F' = (3/8)*F

= (3/8)* 5.6 ( F= 5.6N)

= 2.1 N

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