Question

Charges $-q$ and $+q$ located at $\text{A}$ and $\text{B}$ respectively and $\text{OP}$ is perpendicular to line $\text{AB}$. A charge $Q$ is placed at $\text{P}$, where $\text{OP}=y$ $(y>>2a)$. The charge $Q$ experiences an electrostatics force $F$. If $Q$ is now moved along the equatorial line to $P^{\prime }$ such that $O{P}^{\prime }=\frac{y}{3}$, the force experienced by charge $Q$ will be close to
$(\text{Assume}\frac{y}{2}>>2a)$

• A. $F/3$
• B. $F/27$
• C. $9F$
• D. $27F$

Answer 1

The correct option is D $27F$

For $y>>2a$, net electric field due to a dipole at an equitorial point is given by

$E=\frac{k|\overrightarrow{p}|}{x^3}=\frac{2kql}{x^3}$

Here, $OP=x=y;\text{and}l=a$

So,
$E=\frac{2kqa}{y^3}$

Further, force on charge $Q$

$F=QE=\frac{2kQqa}{y^3}.......\left(1\right)$

If $Q$ is moved to $O{P}^{\prime }=x=y/3$, then

$E^{\prime }=\frac{2kql}{(y/3{)}^3}=27\times \frac{2kqa}{y^3}$

So, force on charge $\left(Q\right)$

$F^{\prime }=Q{E}^{\prime }=27\times \frac{2kQqa}{y^3}$

from eq.(1)

$\Rightarrow F^{\prime }=27F$

So, option $\left(d\right)$ is the correct answer.