Question

Two particles having charges $q_1$ and $q_2$ when kept at a certain distance, exert force $F$ on each other. If distance is reduced to half, force between them becomes :

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

Answer 1

The correct option is C $4F$

Answer is C.

Coulomb's law describes the magnitude of the electrostatic force between two electric charges. The Coulomb's law formula is given as follows.

The force between $2$ charges $q_1$ and $q_2$ is given by the formula $F=k\frac{q_1\times q_2}{r^2}$.

Where, $q_1$: Charge of object $1$

$q_2$: Charge of object $2$

r: Distance between the two objects

F: Force between the two objects. A positive force implies a repulsive interaction, while a negative force implies an attractive interaction

k: Coulomb Constant.

In this case, the distance between $2$ charges is halved. So the new force becomes $F_{new}=k\frac{q_1\times q_2}{{\left(\frac{r}{2}\right)}^2}=\frac{4\times k\times q_1\times q_2}{r^2}=4F$.

Hence, the force between the charges becomes $4F$.