How to answer 3and4 question

How to answer 3and4 question p.c.c.aEavuy by Pm.d-twu 2 -ana.tlt.ac.za an azøanaün;} Pl'na •eu_tlUå ranttéu

Dear Student,

Q3 (I)
Assuming all are of positive magnitude.
The following diagram shows the direction of all the 3 forces experienced by charge q₁

                                           

From diagram: R₁₂=R₁₄=a  and R₁₃=$\sqrt{2}a$ ​
or,

        ​^{$F_{12 }= \frac{K{q}_1{q}_2}{{R_{12}}^2}=\frac{K{q}_1{q}_2}{a^2}$  , 
       $F_{13}= \frac{K{q}_1{q}_3}{{R_{13}}^2}=\frac{K{q}_1{q}_3}{2a^2}$ 
and $F_{14 }= \frac{K{q}_1{q}_4}{{R_{14}}^2}=\frac{K{q}_1{q}_4}{a^2}$}

K=$\frac{1}{4\pi {\epsilon }_0}=9x{10}^9$

Thus the net force experienced by charge q₁ will be resultant of F₁₂ , F₁₃ and F₁₄ 
i.e. F= F₁₂ cos45⁰+ F₁₃ + F₁₄ cos45⁰

$$\begin{gathered} F=\left(\frac{K{q}_1}{a^2}\right)\left(\frac{q_2}{\sqrt{2}}+\frac{q_3}{2}+\frac{q_4}{\sqrt{2}}\right) \\ Here q_1 = q_2 = q_3 = q_4 =q \\ So \\ F=\left(\frac{Kq}{a^2}\right)\left(\frac{q}{\sqrt{2}}+\frac{q}{2}+\frac{q}{\sqrt{2}}\right) =\left(\frac{K{q}^4}{a^2}\right)\left(\frac{2+\sqrt{2}+2}{2\sqrt{2}}\right) =\left(\frac{K{q}^4}{a^2}\right)\left(\frac{4+\sqrt{2}}{2\sqrt{2}}\right) \end{gathered}$$ 
(II)
At centre forces due to all charges are equal in opposite in direction so cancels out. Hence net force is Zero.

Please post different questions in different thread

Regards.