Question 687: Electrostats & Capacitance

Question

Two spherical conductors $$B$$ and $$C$$ having equal radii and carrying equal charges on them repel each other with a force $$F$$ when kept apart at some distance. A third spherical conductor having same radius as that $$B$$ but uncharged is brought in contact with $$B,$$ then brought in correct with $$C$$ and finally removed away from both. The new force of repulsion between $$B$$ and $$C$$ is

Accepted Answer

$$F \propto {{{Q_A}{Q_C}} \over {{x^2}}}$$
$$x$$ is distance  between the spheres. After first operation charge on $$B$$ is halved i.e $${Q \over 2}.$$
and charge on third sphere becomes $${Q \over 2}.$$ Now it is touched to $$C$$, charge then equally
distributes them selves to make potential same, hence charge on $$C$$ becomes
$$\left( {Q + {Q \over 2}} ight){1 \over 2} = {{3Q} \over 4}.$$
$$	herefore$$ $${F_{new}} \propto {{{Q_C}{Q_B}} \over {{x^2}}}$$
$$ = {{\left( {{{3Q} \over 4}} ight)\left( {{Q \over 2}} ight)} \over {{x^2}}}$$
$$ = {3 \over 8}{{{Q^2}} \over {{x^2}}}$$
or, $${F_{new}} = {3 \over 8}F$$

Answer

$$F/8$$

Answer

$$3$$ $$F/4$$

Answer

$$F/4$$

Answer

$$3$$ $$F/8$$

Electrostats & Capacitance

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