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(D) Let $q$ be the magnitude of the two identical point charges placed at $x=0$ and $x=l$.The electric field at a point $P$ at a distance $x$ from the

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(D) Let $q$ be the magnitude of the two identical point charges placed at $x=0$ and $x=l$.The electric field at a point $P$ at a distance $x$ from the first charge (at $x=0$) is the vector sum of the fields due to both charges.The field due to the first charge is $E_1 = \frac{kq}{x^2}$ (directed away from the charge).The field due to the second charge (at $x=l$) is $E_2 = \frac{kq}{(l-x)^2}$ (directed towards the first charge).Taking the direction away from the first charge as positive,the net electric field $E$ at point $P$ is:$E = E_1 - E_2 = kq \left[ \frac{1}{x^2} - \frac{1}{(l-x)^2} ight]$As $x 	o 0$,$E 	o +\infty$.As $x 	o l$,$E 	o -\infty$.At the midpoint $x = l/2$,$E = kq \left[ \frac{1}{(l/2)^2} - \frac{1}{(l/2)^2} ight] = 0$.The curve that shows $E$ starting from positive infinity,passing through zero at $x = l/2$,and approaching negative infinity as $x$ approaches $l$ is represented by option $(D)$.

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