Question

Two point charges a & b whose magnitudes are same positioned at a certain distance from each other, a is at origin. Graph is drawn between electric field strength at points between a & b and distance x from a. E is taken positive if it along the line joining from a to b. Find the graph it can be decided that

Answer 1

Electrical field at a distance $x$ from origin $\left(0<x<x_0\right)$

is given by:

$E=\frac{k{q}_1}{x}\hat{i}+\frac{k{q}_2}{x_0-x}\left(\hat{i}\right)=\left(\frac{k{q}_1}{x}-\frac{k{q}_2}{x_0-x}\right)\hat{i}$

Where $q_1$ is charge on a

Where $q_2$ is charge on b

from graph

As $x\to 0E\to \infty$. this implies that $q_1$ is $+ve$

As,

$\begin{array}{c}x\to x_0{E}_0\to \infty .\\ x_0-x>0,k>0\end{array}$

for $E$ to tend $+\infty$ must be $-ve$ so as $-\frac{k{q}_2}{x_0-x}\to \infty$

Therefore $q_1$ is $+ve$ and $q_2$ is $-ve$