Question

[Honors] Two charged objects with an equal charge of Q separated by a distance r attract each other with a certain force. If the charges on both objects are doubled and the separation is halved, the force between them?(A) 4 times greater (B) 2 times greater (C) 4 times less(D) 16 times greater

Answer 1

The formula for force between charge particles is.

$\begin{gathered} F=k\frac{q1q2}{r^2} \\ Where,\text{ } \\ K=Constant=9\times10^9\frac{Nm^2}{C^2} \\ q1=charge \\ q2=charge \\ r=distance\text{ }between\text{ }charge\text{ }paritlces \end{gathered}$

When charges are r distance apart then force between them is,

$F=K\frac{Q^2}{r^2}$

When charges are doubled and separation is halved then the new force between them is,

$F^{\prime}=K\frac{(2Q)(2Q)}{(\frac{r}{2})^2}=K\frac{4Q^2}{(\frac{r^2}{4})}=K\frac{16Q^2}{r^2}=16F$

So the new force between them is 16 times greater as compare to before.

So option D is correct option.