Question

The force between two charges $2\mu C$ and $4\mu C$ is $24N$, when they are separated by a certain distance in free space. The forces if, $\left(a\right)$ distance between them is doubled and $\left(b\right)$ distance is halved are :

• A. $16N,80N$
• B. $8N,72N$
• C. $6N,96N$
• D. $10N,68N$

Answer 1

Answer: (C)

$6N,96N$

We know $F=\frac{k{q}_1{q}_2}{r^2}=24N$
a) When $r$ is doubled

$F=\frac{k{q}_1{q}_2}{(2r{)}^2}=\frac{k{q}_1{q}_2}{4r^2}=\frac{24}{4}=6N$

b) When $r$ is halved

$F=\frac{k{q}_1{q}_2}{(\frac{r}{2}{)}^2}=\frac{4k{q}_1{q}_2}{r^2}=4\times 24=96N$