Question

A given charge situated at a certain distance from an electric dipole in the end-on
position, experiences a force F. If the distance of the charge is doubled, the force acting on the charge will be :
A. ${\displaystyle \frac{F}{8}}$
B. ${\displaystyle \frac{F}{4}}$
C. ${\displaystyle \frac{F}{2}}$
D. $2F$

Answer 1

Hint:Use the formula for the electric field at a point on the axis of the dipole. Then use the formula for the electric force acting on a charge placed at the point, due to the electric field at that point and find the force on the charge due to dipole.

Formula used:
$E={\displaystyle \frac{2kP}{r^3}}$
$F=qE$

Complete step by step answer:
It is given that a charge is situated at a certain distance from an electric dipole in the end-on
position. This means that the charge is placed on the axis of the dipole. (The axis of a dipole is the line joining the two charge of the dipole)

Suppose the charge is placed at distance r from the midpoint of the dipole.

The electric field at a point (say point P) at a distance r from the dipole on its axis is given as
$E={\displaystyle \frac{2kP}{r^3}}$ , where k is the coulomb’s constant and P is the dipole moment.

When a charge q is placed at a point where the electric field is E, an electric force is exerted on the charge. The magnitude of the electric force is given as $F=qE$.

Therefore, the charge placed at point P will experience a force $F=qE=q\left({\displaystyle \frac{2kP}{r^3}}\right)={\displaystyle \frac{2qkP}{r^3}}$ …. (i).

Now, it said that the distance of the charge from the dipole is doubled. That means the new distance of the charge form the dipole is ‘2r’.

Therefore, the force acting on the charge at this position is equal to $F^{\prime }={\displaystyle \frac{2qkP}{{(2r)}^3}}$.
$\Rightarrow F^{\prime }={\displaystyle \frac{2qkP}{8r^3}}$ … (ii).

Now, divide (ii) by (i).
$\Rightarrow {\displaystyle \frac{F^{\prime }}{F}}={\displaystyle \frac{{\displaystyle \frac{2qkP}{8r^3}}}{{\displaystyle \frac{2qkP}{r^3}}}}$
$\Rightarrow {\displaystyle \frac{F^{\prime }}{F}}={\displaystyle \frac{1}{8}}$
$\Rightarrow F^{\prime }={\displaystyle \frac{F}{8}}$.

This means that if the distance of the charge is doubled, the force acting on the charge will be ${\displaystyle \frac{F}{8}}$.

Hence, the correct option is A.

Note:Note that the formula for the electric field at a point on the axis of the dipole is valid only when the distance of the charge from the dipole is much larger than the distance between the two charges of the dipole.
i.e. r >>> a, where a is the distance between the two charges on the dipole.