Two equal positive point charges are separated by a distance $2 a$ The distance of a point from the centre of the line joining two charges on the equatorial line (perpendicular bisector) at which force experienced by a test charge $q_0$ becomes maximum is $\frac{a}{\sqrt{x}}$ The value of $x$ is ______
Correct Answer: 2
Solution and Explanation
The correct answer is 2
\(F=\frac{2Kqq_0x}{(x^2+a^2)^{3/2}}\)
For F to be maximum
\(\frac{dF}{dx}=0\)
\(x=\frac{a}{\sqrt2}\)
So, the correct answer is \(\frac{a}{\sqrt2}\)
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Concepts Used:
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The motion in a straight line is an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. It is nothing but linear motion.
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