Question

Two electric dipoles of moment $p$ and $64\,p$ are placed in opposite directions on a line at a distance of $25\,cm$. The electric field will be zero at which point between the dipoles whose distance from the dipole of moment $p$?

Answer 1

Use the formula for electric field due to a dipole to find the electric field. The electric field due to the dipole will be zero when for both the dipoles the electric field is zero at that point.

Formula used:
The electric field at a point due to a dipole is given by,
$\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}\hat r$
where $p$ is the dipole moment,$r$ is the distance from the dipole and ${\varepsilon _0}$ is the electric permittivity of the medium is the unit vector along $r$ from the dipole.

Complete step by step answer:
We have here two dipoles that are kept in the same line opposite to each other. We have to find the distance from the dipole where the field is zero. Now the electric field will be zero where the strength of the electric field for both the dipoles will be zero. Then only the net electric field due to the dipoles will be zero. So, we know that the electric field due to a dipole is given by, ,
$\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{r^3}}}\hat r$

So, let’s assume that the electric field is zero at a distance $x\,cm$ from the dipole $p$.So, the electric field due to the dipole $p$ is,
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{x^3}}}$ [taking only the magnitude]
Now the distance between the dipole is $25\,cm$. So, the point at which the electric field is zero is $(25 - x)\,cm$ from the other dipole of strength $64\,p$.
So, the electric field due to the dipole $64p$ is,
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2(64p)}}{{{{(25 - x)}^3}}}$
Now, this two field are equal at that point so we can write,
$\dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2p}}{{{x^3}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{2(64p)}}{{{{(25 - x)}^3}}}$
Or, $\dfrac{1}{{{x^3}}} = \dfrac{{64}}{{{{(25 - x)}^3}}}$
Upon simplifying we have,
$x = \dfrac{{25 - x}}{4}$
$\Rightarrow 5x = 25$
$\therefore x = 5$

So, the electric field will be zero at $5\,cm$ away from the dipole of strength $p$.

Note: An electric dipole is defined as a couple of opposite charges $q$ and $–q$ separated by a distance $d$. By default, the direction of electric dipoles in space is always from negative charge $-q$ to positive charge $q$. The electric field due to dipole at large distance means distance larger than the dimension of dipole is the expression we used. The general expression for electric field due to a dipole is given by, $\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{3(\vec p.\hat r)\hat r - \vec p}}{{{r^3}}}$