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Question

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



A
53 V/μm and in the negative x direction
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B
53 V/μm and in the positive x direction
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C
109 V/μm and in the negative x direction
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D
109 V/μm and in the positive x direction
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Solution

The correct option is C $$\dfrac{10}{9}\ V/\mu m$$ and in the positive x direction
Given $$V(x)=\dfrac{20}{x^{2}-4}$$
We know, $$ E=-\dfrac{dV}{dx}$$
$$\Rightarrow E=-\dfrac{d}{dx}(\dfrac{20}{x^{2}-4})$$
$$\Rightarrow E=-20\dfrac{-1(2x)}{(x^{2}-4)^{2}}$$
$$\Rightarrow E=\dfrac{40x}{(x^{2}-4)^{2}}$$
$$\Rightarrow E=\dfrac{40\times4\ \mu m}{12\times12}$$
$$\Rightarrow E=\dfrac{10}{9} \ V/ \mu m$$

Physics
NCERT
Standard XII

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