Question

If the distance between two equal charges is reduced to half and the magnitude of charges is also decreased to half, then the force between them will be:
(a) Remain same
(b) Decreased to half
(c) Increased to double
(d) Becomes four time​

Answer 1

Given:

The distance between two equal charges is reduced to half and the magnitude of charges is also decreased to half.

To find:

How the force will change?

Calculation:

Initial force is :

$F = \dfrac{k {q}^{2} }{ {r}^{2} }$

• Now, each charge is q/2 and distance is r/2.

Final force is:

$F2= \dfrac{k {( \dfrac{q}{2}) }^{2} }{ {( \dfrac{r}{2}) }^{2} }$

$\implies F2= \dfrac{4}{4} \times \dfrac{k {q}^{2} }{{r}^{2} }$

$\implies F2= \dfrac{k {q}^{2} }{{r}^{2} }$

$\implies F2= F$

So, the final force remains the same as initial force.

Option a) is correct ✔️

Answer 2

a.is correct

Explanation:

there is no change in force