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Question

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


A
5/3 Volt/m and in the ve x direction
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B
5/3Volt/m and in the +ve x direction
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C
10/9 Volt/m and in the ve x direction
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D
10/9 Volt/m and in the +ve x direction
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Solution

The correct option is A $$10/9$$ Volt/m and in the $$+ve$$ x direction
The relation between Electric field and electrostatic potential is given by 

$$\displaystyle \vec{E} = -\nabla V$$

In one dimension, $$\vec{E} = \displaystyle -\frac{dV}{dx}$$

Given $$V = \displaystyle \frac{20}{x^2-4}$$

Thus, $$E = -\displaystyle \frac{dV}{dx} = \frac{40x}{(x^2-4)^2}$$

Thus Field at $$x = 4\textrm{ m}$$ is given by $$E = \displaystyle\frac{40\times 4}{(4^2-4)^2} = +\frac{10}{9} \textrm{ V/m}$$

The field is in the positive $$x$$ direction

Physics
NCERT
Standard XII

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