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Two point charges a and b whose magnitudes are same positioned at a certain distance along the positive x-axis from each other . a is at origin and b is at a distance $\;x_{0}\;$ from a . Graph is drawn between electrical field strength and distance x from a . E is taken positive if it is along the line joining from a to b . Then

$(a)\;a\;is\;+ve,b\;is\;-ve\qquad(b)\;a\;and\;b\;both\;are\;+ve\qquad(c)\;both\;are\;-ve\qquad(d)\;a\;is\;-ve\;,b\;is\;+ve$

Answer : (a) a is +ve , b is -ve
Explanation :
Electrical field at a distance x from origin $\;(0 < x < x_{0})\;$ is given by :
$E=\large\frac{k\;q_{1}}{x} \hat{i} + \large\frac{k\;q_{2}}{x_{0}-x} (-\hat{i})=(\large\frac{k\;q_{1}}{x}-\large\frac{k\;q_{2}}{x_{0}-x})\;\hat{i}$
Where $\;q_{1}\;$ is charge on a
Where $\;q_{2}\;$ is charge on b
From graph
As $\;x \;\to\;0\quad \;E\;\to\;+\infty\;.$ this implies that $\;q_{1}\;$ is +ve
As $\;x\;\to\;x_{0} \quad \;E_{0}\;\to\;+\infty\;.$
$x_{0}-x > 0 \quad \;, k > 0$
For E to tend $\;+\infty\;q_{2}\;$ must be $\;-ve\;$ so as $\;-\large\frac{k\;q_{2}}{x_{0}-x}\;\to\;-\infty$
Therefore $\;q_{1}\;$ is +ve and $\;q_{2}\;$ is -ve .
edited Feb 12, 2014 by yamini.v