Question 665: Electrostats & Capacitance

Question

The potential at a point $$x$$ (measured in $$\mu \,m$$) due to some charges situated on the $$x$$-axis is given by $$V\left( x \right) = 20/\left( {{x^2} - 4} \right)$$ volt
The electric field $$E$$ at $$x = 4\,\mu \,m$$ is given by

Accepted Answer

Here, $$V\left( x ight) = {{20} \over {{x^2} - 4}}volt$$
We know that $$E =  - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} ight)$$
or, $$E =  + {{40x} \over {{{\left( {{x^2} - 4} ight)}^2}}}$$
At $$x = 4\mu m,$$
$$E =  + {{40 	imes 4} \over {{{\left( {{4^2} - 4} ight)}^2}}}$$
$$ =  + {{160} \over {144}} =  + {{10} \over 9}volt/\mu m.$$
Positive sign indicates that $$\overrightarrow E $$ is in $$+ve$$ $$x$$-direction.

Answer

$$(10/9)$$ volt / $$\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction

Answer

$$\left( {5/3} \right)$$ volt/ $$\mu $$ $$m$$ and in the $$-ve$$ $$x$$ direction

Answer

$$\left( {5/3} \right)$$ volt/$$\mu $$ $$m$$ and in the $$+ve$$ $$x$$ direction

Answer

$$\left( {10/9} \right)$$ volt/ $$\mu \,m$$ and in the $$-ve$$ $$x$$ direction

Electrostats & Capacitance

The potential at a point $x (measured in \mu \,m) due to some charges situated on the x-axis is given by V\left( x \right) = 20/\left( {{x^2} - 4} \right) volt The electric field E at x = 4\,\mu \,m$ is given by

Choose the correct answer: