The net force on the point charge +Q is \(1.68 k \dfrac {Q^2}{d^2}\;\rm N\).
How do you calculate the force?
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 \(\sqrt{2}d\). 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.
\(F'' = k \dfrac {(+2Q )( +Q)}{d^2}\)
\(F'' = k\dfrac {2Q^2}{d^2} \;\rm N\)
The force between the point charge +Q and -Q is given below.
\(F' = k\dfrac {(+Q)(-Q)}{(\sqrt{2}d)^2}\)
\(F' = k\dfrac {-Q^2}{2d^2} \;\rm N\)
The net force along the x-axis is given below.
\(F_x = F' sine 45^\circ\)
\(F_x = k\dfrac {-Q^2}{2d^2} \times \dfrac {1}{\sqrt{2} }\)
The net force along the y-axis is given below.
\(F_y = F'' + F'cos 45^\circ\)
\(F_y = k\dfrac {2Q^2}{d^2} + k\dfrac {(-Q^2)}{2d^2} \times \dfrac {1}{\sqrt{2} }\)
\(F_y = k\dfrac {Q^2}{d^2}({2 -\dfrac {1}{2\sqrt{2}} )\)
The net force on the point charge +Q is given below.
\(F = \sqrt{F_x^2 + F_y^2}\)
\(F = \sqrt{(\dfrac {-kQ^2}{2\sqrt{2}d^2} )^2 + (\dfrac {kQ^2}{d^2}(2 - \dfrac {1}{2\sqrt{2} }))^2\)
\(F = \dfrac {kQ^2}{d^2} \sqrt{\dfrac {1}{8} + (2-\dfrac {1}{2\sqrt{2}})^2 }\)
\(F = 1.68 k\dfrac {Q^2}{d^2}\;\rm N\)
Hence we can conclude that the net force on the point charge +Q is \(1.68 k \dfrac {Q^2}{d^2}\;\rm N\).
To know more about the net force, follow the link given below.
https://brainly.com/question/16985000.
