Answer to Question #203060 in Electric Circuits for Nabeel khan

Question

Two charges 2C and -2C are at 0.01m. Calculate the electric field intensity 0.3m in equatorial plane.

Answer 1

Let us consider two equal and opposite charges $q$ and $-q$ at points A and B separated by a distance $2d$ as shown in the figure. We have to find the electric field at an equatorial point $P$ distant $x$ from the center $O$ .

From $\Delta AOP$ and $\Delta BOP$ ,

$AP=BP=\sqrt{x^2+d^2}$

The electric field at P due to $q$ is

$\vec{E}_a=\frac{1}{4\pi \epsilon_0}\frac{q}{x^2+d^2}$ directed along AP.

The electric field at P due to $-q$ is

$\vec{E}_b=\frac{1}{4\pi \epsilon_0}\frac{q}{x^2+d^2}$ directed along PB.

Note that the electric fields $\vec{E}_a$ and $\vec{E}_b$ have equal magnitude i.e,

$|\vec{E}_a|=|\vec{E}_b|=E(say)$

To find the resultant electric field at $P$ we will resolve $\vec{E}_a$ and $\vec{E}_b$ in horizontal and vertical components as shown below.

The vertical components $E_a\sin\theta$ and $E_b\sin\theta$ will cancel each other (since $|\vec{E}_a|=|\vec{E}_b|$ ).

The horizontal components $E_a\cos\theta$ and $E_b\cos\theta$ will add up to give the resultant electric field.

$\therefore$ Resultant electric field at $P$ is

$E_r=E_a\cos\theta+E_b\cos\theta\\ \Rightarrow E_r=2E\cos\theta\\ \Rightarrow E_r=2\frac{1}{4\pi\epsilon_0}\frac{q}{x^2+d^2}.\frac{d}{\sqrt{x^2+d^2}}\\ \Rightarrow \boxed{ E_r=\frac{1}{4\pi\epsilon_0}\frac{q.2d}{(x^2+d^2)^{3/2}}}$ directed along AB.

Given, $q=2C, 2d=0.01\ m, x=0.3\ m$

We know $\frac{1}{4\pi\epsilon_0}=9\times 10^9 N.m^2.C^{-2}$

$\therefore E_r=9\times 10^9\times \frac{2\times 0.01}{(0.3^2+0.005^2)^{3/2}}\ N/C\\ E_r=6.7\times 10^9\ N/C$

Answer: The electric filed intensity at 0.3m in equatorial plane is $6.7\times 10^9 \ N/C$