Question
The potential at a point $$x$$ (measured in $$\mu m$$ ) due to some charges situated on the $$x$$-axis is given by $$V\left( x \right) = \frac{{20}}{{\left( {{x^2} - 4} \right)}}volt.$$ The electric field $$E$$ at $$x = 4\,\mu m$$ is given by
A.
$$\left( {\frac{{10}}{9}} \right)volt/\mu m$$ in the $$+ve$$ $$x$$ direction
B.
$$\left( {\frac{{5}}{3}} \right)volt/\mu m$$ in the $$-ve$$ $$x$$ direction
C.
$$\left( {\frac{{5}}{3}} \right)volt/\mu m$$ in the $$+ve$$ $$x$$ direction
D.
$$\left( {\frac{{10}}{9}} \right)volt/\mu m$$ in the $$-ve$$ $$x$$ direction
Answer :
$$\left( {\frac{{10}}{9}} \right)volt/\mu m$$ in the $$+ve$$ $$x$$ direction
Solution :
Here, $$V\left( x \right) = \frac{{20}}{{{x^2} - 4}}volt$$
We know that $$E = - \frac{{dV}}{{dx}} = - \frac{d}{{dx}}\left( {\frac{{20}}{{{x^2} - 4}}} \right)$$
$$\eqalign{
& {\text{or,}}\,\,E = + \frac{{40x}}{{{{\left( {{x^2} - 4} \right)}^2}}} \cr
& {\text{At}}\,\,x = 4\,\mu m, \cr
& E = + \frac{{40 \times 4}}{{{{\left( {{4^2} - 4} \right)}^2}}} = + \frac{{160}}{{144}} = + \frac{{10}}{9}\,volt/\mu m. \cr} $$
Positive sign indicates that $${\vec E}$$ is in $$+ve$$ $$x$$-direction.