Q. The potential at a point $x$(measured in $\mu$m) due to some charges situated on the $x$-axis is given by $V(x) = 20/(x^{2} - 4)\ volt$ The electric field E at $x$=4$\ \mu$m is given by
A) $\frac{5}{3}V\mu^{- 1}m^{- 1}\ and\ in\ the - ve\ x\ direction$
B) $\frac{5}{3}V\mu^{- 1}m^{- 1}\ and\ in\ the + ve\ x\ direction$
C) $\frac{10}{9}V\mu^{- 1}m^{- 1}\ and\ in\ the - ve\ x\ direction$
D) $\frac{10}{9}V\mu^{- 1}m^{- 1}\ and\ in\ the + ve\ x\ direction$
Explanation
(d) $\mathbf{E} = - \frac{\partial V}{\partial x}\widehat{\mathbf{I}} - \frac{\partial V}{\partial y}\widehat{\mathbf{J}} - \frac{\partial V}{\partial z}\widehat{\mathbf{k}}\ $
$\Longrightarrow E_{x} = - \frac{\partial V}{\partial x} = - \frac{d}{dx}\left\lbrack \frac{20}{x^{2} - 4} \right\rbrack = - \frac{40x}{\left( x^{2} - 4 \right)^{2}}$
$\Longrightarrow E_{x}at\ x = 4\mu m = \frac{10}{9}V\mu^{- 1}m^{- 1}$
And is long positive $x$-direction
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.