Question 609: Electrostats & Capacitance

Question

Two identical conducting spheres A and B,  carry equal charge. They are separated by a distance much larger than their diameters,  and the force between theis F.  A third identical conducting sphere, C,  is uncharged. Sphere C is first touhed to A, then to B, and then removed. As a result, the force between A and B would be equal to :

Accepted Answer

Let, change of A and B = q
$$	herefore\,\,\,$$ Force between them, F = $${{k 	imes q 	imes q} \over {{r^2}}} = {{k{q^2}} \over {{r^2}}}$$
When C touched with A then charge of A. Will fl;ow to C and divide into half parts.
$$	herefore\,\,\,$$ charge of A and C ,
q
A
= q
B
= $${q \over 2}$$
Then C touched with B, then charge on B,
q
B
= $${{{q \over 2} + q} \over 2} = {{3q} \over 4}$$
$$	herefore\,\,\,$$ Force between A and B,
F' = $${{k 	imes {q \over 2} 	imes {{3q} \over 4}} \over {{r^2}}}$$
= $${{k 	imes 3{q^2}} \over {8{r^2}}}$$
= $${3 \over 8} 	imes {{k{q^2}} \over {{r^2}}}$$
= $${3 \over 8}\,F$$

Answer

F

Answer

$${{3F} \over 4}$$

Answer

$${{3F} \over 8}$$

Answer

$${{F} \over 2}$$

Electrostats & Capacitance

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