Two identical charged conducting spheres A and B have their centres...

Two identical charged conducting spheres

A

and

B

have their centres separated by a certain distance. Charge on each sphere is q and the force of repulsion between them is

F

. A third identical uncharged conducting sphere is brought in contact with sphere A first and then with B and finally removed from both. New force of repulsion between spheres

A

and

B

(Radii of

A

and

B

are negligible compared to the distance of separation so that for calculating force between them they can be considered as point charges) is best given as : Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each

Two identical charged conducting spheres A and B have their centres separated by a certain distance. Charge on each

Options

A $\frac{3 F}{5}$
B $\frac{2 F}{3}$
C $\frac{F}{2}$
D $\frac{3 F}{8}$

Correct answer

D. $\frac{3 F}{8}$

Step-by-step solution

\mathrm{F}=\frac{\mathrm{kQ}^2}{\mathrm{r}^2}

on touching;

F^{\prime}=\frac{k(Q / 2)(3 Q / 4)}{r^2}F^{\prime}=\frac{3 F}{8}

After touching \rightarrow(

\mathrm{A}

) \&(

\mathrm{C}

) \rightarrow

Total charge \rightarrow \mathrm{q}+0=\mathrm{q} \mathrm{Q}_{\mathrm{A}}^{\prime}=\frac{\mathrm{q}}{2}, \quad \mathrm{Q}_{\mathrm{C}}^{\prime}=\frac{\mathrm{q}}{2}

(B) & (C)

\begin{aligned} & \text{ Total charge }=q+\frac{q}{2}=\frac{3}{2} q \\ & Q_B^{\prime}=\frac{3}{4} q \quad Q_C^{\prime}=\frac{3}{4} q \end{aligned}

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