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:
Motion in a straight line
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.
Types of Linear Motion:
Linear motion is also known as the Rectilinear Motion which are of two types:
- Uniform linear motion with constant velocity or zero acceleration: If a body travels in a straight line by covering an equal amount of distance in an equal interval of time then it is said to have uniform motion.
- Non-Uniform linear motion with variable velocity or non-zero acceleration: Not like the uniform acceleration, the body is said to have a non-uniform motion when the velocity of a body changes by unequal amounts in equal intervals of time. The rate of change of its velocity changes at different points of time during its movement.