Question

In the following figure, the electric field on $y$ -axis will be maximum at $y=$

• A. -$\frac{d}{2}$
• B. -$\frac{d}{3}$
• C. $\frac{d}{\sqrt{2}}$
• D. $\frac{d}{\sqrt{3}}$

Answer 1

The correct option is A -$\frac{d}{2}$

$\frac{dE}{dY}=0$

or, when at $P$ is loaded at the perpendicular bisector of a dipole

$E=\frac{P}{{(n^2+l^2)}^{3/2}}r=Y$

$\frac{dE}{dY}=\frac{P}{dy{\left(Y^2+l^2\right)}^{3/2}}=0$

$=P{\frac{d}{dy}(Y^2+l^2)}^{-3/2}=0$

or, $P\times \frac{-3}{2}{(y^2+d^2)}^{-3/2-1}=0$

or, $P\times (-\frac{3}{2}){(y^2+d^2)}^{-\frac{5}{2}}=0$

where $l=d$

Now,

$P(-\frac{3}{2}){(y^2+d^2)}^{\frac{-5}{2}}=0$

or, $d^2+y^2=0$

or, $d=-y$

or, $y=-\frac{d}{2}$