Question

the potential at a point X due to some charges situated in x axis is given by V = 20/(x^2-4).Find thr electric field at x=4micro metre

Answer 1

Final Answer : E = 10^(-5) V/m
Direction: Positive direction of x-axis

For Calculation see pic

Answer 2

Explanation :

Potential at X is, $V=\dfrac{20}{x^2-4}$

The relation between the electric field and the electric potential is given by :

$E=\dfrac{-dV}{dx}$

$E=\dfrac{d}{dx}(-\dfrac{20}{x^2-4})$

$E=\dfrac{40x}{(x^2-4)^2}$

At $x=4\times 10^{-6}\ m$

$E=\dfrac{-40\times 4\times 10^{-6}}{[(4\times 10^{-6})^2-4]^2}$

The denominator becomes very small so neglecting x here.

So, $E=\dfrac{-40\times 4\times 10^{-6}}{(-4)^2}$  

$E=-0.00001\ V/m$

$E=10^{-5}\ V/m$

Hence, this is the required solution.