Question

The electric force between two identical charge particle having separation $r$ between them is $F$. If separation between them is increased twice to initial separation keeping charge constant, then new electrical force between them is

• A. $F$
• B. $4F$
• C. $\frac{F}{4}$
• D. $\frac{F}{2}$

Answer 1

The correct option is C $\frac{F}{4}$

Let the two identical charge particles of charge $Q$ be separated at a distance $r$. According to Coulomb’s law, electric force acting between them is given by:
$F=k\frac{Q^2}{r^2}$
$\Rightarrow F\propto \frac{1}{r^2}$
Let $F\text{'}$ be the new electric force acting between the charges when separation between them $\left(r^{\prime }\right)$ is increased twice to the initial separation $\left(r\right)$.
$\therefore \frac{F^{\prime }}{F}={\left(\frac{r}{r^{\prime }}\right)}^2={\left(\frac{r}{2r}\right)}^2=\frac{1}{4}$
$\Rightarrow F^{\prime }=\frac{F}{4}$
Hence, the correct answer is option (c).