Question

Two equal point charges of same sign are fixed on y axis, to the two sides of the origin equidistant from it, distance being d. A third charges move along x axis. Find the distance distance from either of the two fixed charges, when force is maximum

Answer 1

Given:

Two equal point charges of same sign are fixed on y axis, to the two sides of the origin equidistant from it, distance being d. A third charge moves along x axis.

To find:

Distance of the third charge from either of the 2 charge.

Calculation:

Let the third charge be located x distance from origin.

Hence the components of the forces can be resolved, such that the sine components of the force will get cancelled and only the cos component will remain.

$\therefore \: F \: net= 2F \: \cos( \theta)$

$= > \: F = \dfrac{2kQq}{ {( \sqrt{ {x}^{2} + {d}^{2} }) }^{2} } \: \cos( \theta)$

$= > \: F = \dfrac{2kQq}{ {( \sqrt{ {x}^{2} + {d}^{2} }) }^{2} } \: \times \dfrac{x}{ \sqrt{ {x}^{2} + {d}^{2} } }$

$= > \: F = 2kQq \times \dfrac{x}{ {( {x}^{2} + {y}^{2}) }^{ \frac{3}{2} } }$

This kind of function will have a graph as in the attachment. Please refer to it.

This kind of function have maxima at x = d/√2.

$\boxed{ \sf{F _{max} \: at \: x = \dfrac{d}{ \sqrt{2} } }}$

Hence the required distance will be :

$d_{net} = \sqrt{ {d}^{2} + {( \dfrac{d}{ \sqrt{2} }) }^{2} }$

$= > d_{net} = \sqrt{ \dfrac{3 {d}^{2} }{2} }$

$= > d_{net} = d \sqrt{ \dfrac{3}{2} }$

So , final answer is :

$\boxed{ \sf{d_{net} = d \sqrt{ \dfrac{3}{2} }}}$