Question 439: Electrostats & Capacitance

Question

Two identical metallic spheres $$\mathrm{A}$$ and $$\mathrm{B}$$ when placed at certain distance in air repel each other with a force of $$\mathrm{F}$$. Another identical uncharged sphere $$\mathrm{C}$$ is first placed in contact with $$\mathrm{A}$$ and then in contact with $$\mathrm{B}$$ and finally placed at midpoint between spheres A and B. The force experienced by sphere C will be:

Accepted Answer

Let two identical spheres have charge q. And distance between them = r
$$	herefore$$ Force between the spheres $$(F) = {{k{q^2}} \over {{r^2}}}$$
Now when an identical uncharged sphere C comes in contact with A, charge q on sphere A get's divided equally to both sphere.
So, both sphere A and C have charge $$ = {q \over 2}$$
Now, C get's in contact with B. So their total charge $$\left( {q + {q \over 2}} ight)$$ gets divided equally.
So, charge on both B and C is $$ = {{q + {q \over 2}} \over 2} = {{3q} \over 4}$$
Now, C is place midpoint between A and B.
Repulsion force between A and C,
$${F_{AC}} = {{k\left( {{q \over 2}} ight)\left( {{{3q} \over 4}} ight)} \over {{{\left( {{r \over 2}} ight)}^2}}} = {{4k 	imes 3{q^2}} \over {8{r^2}}}$$
Repulsion force between B and C,
$${F_{BC}} = {{k\left( {{{3q} \over 4}} ight)\left( {{{3q} \over 4}} ight)} \over {{{\left( {{r \over 2}} ight)}^2}}} = {{4k 	imes 9{q^2}} \over {16{r^2}}}$$
$$	herefore$$ Net force on C,
$${F_{net}} = {F_{BC}} - {F_{AC}}$$
$$ = {{4k 	imes 9{q^2}} \over {16{r^2}}} - {{4k 	imes 3{q^2}} \over {8{r^2}}}$$
$$ = {{k{q^2}} \over {{r^2}}}\left( {{9 \over 4} - {3 \over 2}} ight)$$
$$ = {{k{q^2}} \over {{r^2}}} 	imes {3 \over 4}$$
$$ = {3 \over 4}F$$ [as $$F = {{k{q^2}} \over {{r^2}}}$$]

Answer

3F/2

Answer

3F/4

Answer

F

Answer

2F

Electrostats & Capacitance

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