Answer: A symbolic expression for the net force on a third point charge +Q located along the y axis
[tex]F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}[/tex]
Explanation:
Let the force on +Q charge y-axis due to +2Q charge be [tex]F_1[/tex] and force on +Q charge y axis due to -Q charge on x-axis be [tex]F_2[/tex].
Distance between the +2Q charge and +Q charge = d units
Distance between the -Q charge and +Q charge = [tex]\sqrt{2}d[/tex] units
[tex]k_e[/tex]= Coulomb constant
[tex]F_1=k_e\frac{(+2Q)(+Q)}{d^2}=k_e\frac{+2Q^2}{d^2} N[/tex]
[tex]F_2=k_e\frac{(-Q)(+Q)}{(\sqrt{2}d)^2}=k_e\frac{-Q^2}{2d^2} N[/tex]
Net force on +Q charge on y-axis is:
[tex]F_x=F_2sin 45^o=k_e\frac{-Q^2}{2d^2}\times \frac{1}{\sqrt{2}} N[/tex]
[tex]F_y=F_1-F_2cos45^o[/tex]
[tex]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}})[/tex]
[tex]F_N=\sqrt{F_x^2+F_y^2}[/tex]
[tex]|F_N|=|k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}|[/tex]
The net froce on the +Q charge on y-axis is
[tex]F_N=k_e\frac{Q^2}{d^2}\times \sqrt{[4+\frac{1}{4}-\sqrt{2}]}[/tex]
The net force on the point charge +Q is [tex]1.68 k \dfrac {Q^2}{d^2}\;\rm N[/tex].
Given that the charge at the origin is 2Q and the charge along the x-axis is -Q. The distance between the point charge 2Q and -Q is d.
The point charge along the y-axis is +Q. The distance between the point charge 2Q and +Q is d. The distance between the point charge +Q and -Q is [tex]\sqrt{2}d[/tex]. The attachment shows the diagram of point charges along x and y axis.
If k is the coulombs constant, then the force between the point charge 2Q and +Q is given below.
[tex]F'' = k \dfrac {(+2Q )( +Q)}{d^2}[/tex]
[tex]F'' = k\dfrac {2Q^2}{d^2} \;\rm N[/tex]
The force between the point charge +Q and -Q is given below.
[tex]F' = k\dfrac {(+Q)(-Q)}{(\sqrt{2}d)^2}[/tex]
[tex]F' = k\dfrac {-Q^2}{2d^2} \;\rm N[/tex]
The net force along the x-axis is given below.
[tex]F_x = F' sine 45^\circ[/tex]
[tex]F_x = k\dfrac {-Q^2}{2d^2} \times \dfrac {1}{\sqrt{2} }[/tex]
The net force along the y-axis is given below.
[tex]F_y = F'' + F'cos 45^\circ[/tex]
[tex]F_y = k\dfrac {2Q^2}{d^2} + k\dfrac {(-Q^2)}{2d^2} \times \dfrac {1}{\sqrt{2} }[/tex]
[tex]F_y = k\dfrac {Q^2}{d^2}({2 -\dfrac {1}{2\sqrt{2}} )[/tex]
The net force on the point charge +Q is given below.
[tex]F = \sqrt{F_x^2 + F_y^2}[/tex]
[tex]F = \sqrt{(\dfrac {-kQ^2}{2\sqrt{2}d^2} )^2 + (\dfrac {kQ^2}{d^2}(2 - \dfrac {1}{2\sqrt{2} }))^2[/tex]
[tex]F = \dfrac {kQ^2}{d^2} \sqrt{\dfrac {1}{8} + (2-\dfrac {1}{2\sqrt{2}})^2 }[/tex]
[tex]F = 1.68 k\dfrac {Q^2}{d^2}\;\rm N[/tex]
Hence we can conclude that the net force on the point charge +Q is [tex]1.68 k \dfrac {Q^2}{d^2}\;\rm N[/tex].
To know more about the net force, follow the link given below.
https://brainly.com/question/16985000.
