Answer:
Horizontal component = [tex]2.62\times10^{-4}N[/tex]
Vertical component = [tex]1.4\times10^{-3}N[/tex]
Explanation:
Firstly, we shall calculate the angle the diagonals make with the horizontal sides as it will be required to solve both parts of the given question,
[tex]\theta=tan^{-1}(\frac{9}{28})=17.82^o[/tex]
Horizontal component:
The upper left charge does not contribute any force towards the lower left charge. So we shall only be concerned with the negative charges on the right.
Force due to the lower right charge (using Coulomb's law) = [tex]k\frac{Q^2}{D_1^2} = (9\times 10^9)\frac{35^2\times 10^{-18}}{28^2\times100^{-2}}= 1.41\times10^{-4}N[/tex]
The whole force in the above case was in the horizontal direction. Now, to calculate the force due to the upper right charge, we shall have to multiply with [tex]cos\theta[/tex] :
Force = [tex]k\frac{Q^2}{D_1^2+D_2^2}cos\theta=(9\times10^9)\frac{35^2\times10^{-18}}{(28^2\times100^{-2})+(9^2\times100^{-2})}cos(17.82^o)N=1.21\times10^{-4}N[/tex]
Hence, total force = [tex]2.62\times10^{-4}N[/tex] ( directed towards the right, since the nature of forces is attractive)
Vertical component:
This time, the lower right charge does not contribute any force in the vertical component.
Force due to the upper left charge
= [tex]k\frac{Q^2}{D_2^2} =(9\times10^9)\frac{35^2\times10^{-18}}{9^2\times100^{-2}}N=1.36\times10^{-3}N[/tex]
Force due to upper right charge (the [tex]sin\theta[/tex] factor comes to get the vertical component)
= [tex]k\frac{Q^2}{D_1^2+D_2^2} sin\theta=(9\times10^9)\frac{35^2\times10^{-18}}{(28^2\times100^{-2})+(9^2\times100^{-2})}sin(17.82^o)N= 0.039\times10^{-3}N[/tex]
Therefore, total force = [tex]1.399\times10^{-3}N\approx1.4\times10^{-3}N[/tex] (along the bottom direction since the repulsive force of the upper left charge is greater than the attractive force of the upper right)
