JEE MAIN - Physics (2018 - 16th April Morning Slot - No. 11)
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 :
F
$${{3F} \over 4}$$
$${{3F} \over 8}$$
$${{F} \over 2}$$
Explanation
Let, change of A and B = q
$$\therefore\,\,\,$$ Force between them, F = $${{k \times q \times q} \over {{r^2}}} = {{k{q^2}} \over {{r^2}}}$$
When C touched with A then charge of A. Will fl;ow to C and divide into half parts.
$$\therefore\,\,\,$$ charge of A and C ,
qA = qB = $${q \over 2}$$
Then C touched with B, then charge on B,
qB = $${{{q \over 2} + q} \over 2} = {{3q} \over 4}$$
$$\therefore\,\,\,$$ Force between A and B,
F' = $${{k \times {q \over 2} \times {{3q} \over 4}} \over {{r^2}}}$$
= $${{k \times 3{q^2}} \over {8{r^2}}}$$
= $${3 \over 8} \times {{k{q^2}} \over {{r^2}}}$$
= $${3 \over 8}\,F$$
$$\therefore\,\,\,$$ Force between them, F = $${{k \times q \times q} \over {{r^2}}} = {{k{q^2}} \over {{r^2}}}$$
When C touched with A then charge of A. Will fl;ow to C and divide into half parts.
$$\therefore\,\,\,$$ charge of A and C ,
qA = qB = $${q \over 2}$$
Then C touched with B, then charge on B,
qB = $${{{q \over 2} + q} \over 2} = {{3q} \over 4}$$
$$\therefore\,\,\,$$ Force between A and B,
F' = $${{k \times {q \over 2} \times {{3q} \over 4}} \over {{r^2}}}$$
= $${{k \times 3{q^2}} \over {8{r^2}}}$$
= $${3 \over 8} \times {{k{q^2}} \over {{r^2}}}$$
= $${3 \over 8}\,F$$
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