QJEEM07Phb463: JEE Main 2007 Physics - Class 12 JEEM07Phb463 Solution: The potential at a point (measured in ) due to some charges…

Question

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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 \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.

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