Question

A point charge +2Q is at the origin and a point charge −Q is located along the x axis at x = d as in the figure below. Find a symbolic expression for the net force on a third point charge +Q located along the y axis at y = d. (Use the following as necessary: ke, the Coulomb constant, Q, and d.)

Answer 1

Force F1, exerted by the charge +2Q located at the origin over the charge +Q at y =d

F1 =  [ k (+2Q) * (+Q) / (d^2) ] j =  [2KQ^2 / d^2] j

Force F2, exerted by the charge -Q located at x = d over the charge + Q located at y = d

Horizontally: F2,x = [k (-Q)(+Q) / (2d^2) ] cos(45°) i = - [KQ^2 / 2d^2] (√2) / 2 i

=- [KQ^2 (√2) / 4d^2]  i


Vertically: F2,y =  [k (-Q)(+Q) / (2d^2) ] sin(45°) j = - [KQ^2 / 2d^2] (√2) / 2 j =

= - [KQ^2 (√2) / 4d^2]  j


Now sum the respective components

Net force = - [KQ^2 (√2) / 4d^2]  i +  [2KQ^2 / d^2] j - [KQ^2 (√2) / 4d^2]  j

Net force = - [KQ^2 (√2) / 4d^2]  i +  [KQ^2 / d^2](2 - (√2)/4 ) j

Answer: - [KQ^2 (√2) / 4d^2]  i +  [KQ^2 / d^2](2 - (√2)/4 ) j

Answer 2

Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis  

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$

Explanation:

Let the force on +Q charge y-axis due to +2Q charge be $F_1$ and force on +Q charge y axis due to -Q charge on x-axis be $F_2$.

Distance between the +2Q charge and +Q charge = d units

Distance between the -Q charge and +Q charge = $\sqrt{2}d$ units

$k_e$= Coulomb constant

$F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N$

$F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N$

Net force on +Q charge on y-axis is:

$F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N$

$F_y=F_1-F_2cos45^o$

$F_y=(F_1-F_2cos45^o)=(k_e\frac{+2Q^2}{d^2})-(k_e\frac{-Q^2}{2d^2}\frac{1}{\sqrt{2}})$

$F_N=\sqrt{F_x^2+F_y^2}$

$|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|$

The net froce on the +Q charge on y-axis is

$F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}$