JEE MAIN - Physics (2007 - No. 20)
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
The electric field $$E$$ at $$x = 4\,\mu \,m$$ is given by
$$(10/9)$$ volt / $$\mu $$ $$m$$ and in the $$ + ve$$ $$x$$ direction
$$\left( {5/3} \right)$$ volt/ $$\mu $$ $$m$$ and in the $$-ve$$ $$x$$ direction
$$\left( {5/3} \right)$$ volt/$$\mu $$ $$m$$ and in the $$+ve$$ $$x$$ direction
$$\left( {10/9} \right)$$ volt/ $$\mu \,m$$ and in the $$-ve$$ $$x$$ direction
Explanation
Here, $$V\left( x \right) = {{20} \over {{x^2} - 4}}volt$$
We know that $$E = - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} \right)$$
or, $$E = + {{40x} \over {{{\left( {{x^2} - 4} \right)}^2}}}$$
At $$x = 4\mu m,$$
$$E = + {{40 \times 4} \over {{{\left( {{4^2} - 4} \right)}^2}}}$$
$$ = + {{160} \over {144}} = + {{10} \over 9}volt/\mu m.$$
Positive sign indicates that $$\overrightarrow E $$ is in $$+ve$$ $$x$$-direction.
We know that $$E = - {{dV} \over {dx}} = {d \over {dx}}\left( {{{20} \over {{x^2} - 4}}} \right)$$
or, $$E = + {{40x} \over {{{\left( {{x^2} - 4} \right)}^2}}}$$
At $$x = 4\mu m,$$
$$E = + {{40 \times 4} \over {{{\left( {{4^2} - 4} \right)}^2}}}$$
$$ = + {{160} \over {144}} = + {{10} \over 9}volt/\mu m.$$
Positive sign indicates that $$\overrightarrow E $$ is in $$+ve$$ $$x$$-direction.
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