Given is the arrangement of electric charges in the picture below. The values of the point charges are
In the beginning, the area charge is $q_4=0 ~\rm nC$.
The permittivity is $\varepsilon_{\rm r} \varepsilon_0 = 8.854 \cdot 10^{-12} ~\rm As/Vm$
1. Calculate the single forces $\vec{F}_{01}$, $\vec{F}_{02}$, $\vec{F}_{03}$, on the charge $q_0$!
First, calculate the magnitude of the forces, like $\vec{F}_{01}$.
The force $\vec{F}_{01}$ is purely on the $x$-axis and therefore equal to $F_{01,x}$.
\begin{align*}
\vec{F}_{01} = F_{01,x} &= {{1}\over{4\pi\varepsilon}}\cdot {{q_1\cdot q_0}\over{r^2_{01}}} \\
&= {{1}\over{4\pi\cdot 8.854 \cdot 10^{-12} ~\rm As/Vm}}\cdot {{1 \cdot 10^{-9} ~\rm C \cdot 5 \cdot 10^{-9} ~\rm C}\over{(7 \cdot 10^{-3} ~\rm m)^2}} \\
&= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{(As)^2 \cdot Vm}\over{As \cdot m^2}}}
= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{VAs}\over{m}}}
= 917.\;.\!.\!.\! \cdot 10^{-6} {\rm {{Ws}\over{m}}} \\
&= 917.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)}
\end{align*}
Similarly, we get for $\vec{F}_{02}$ and $\vec{F}_{03}$ \begin{align*} \vec{F}_{02} = F_{02,x} &= -1997.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the right)} \\ \vec{F}_{03} = F_{03,y} &= -1123.\;.\!.\!.\! {\rm \mu N} \quad \text{(to the top)} \\ \end{align*}
2. What is the magnitude of the resulting force?
3. Now the charges $q_2=q_3$ are set to 0. The area charge $q_4$ generates a homogenous field of $E_4$. Which value needs $E_4$ to have to get a resulting force of $0 ~\rm N$ on $q_0$?
In the homogenous field the force is calculated by $F = E \cdot q$.
Here, this field has to compensate for the force $\vec{F}_{01}$ from $q_1$ on $q_0$:
\begin{align*}
|\vec{F}_{01}| &= |E_4| \cdot |q_0| \\
\rightarrow E_4 &= {{|\vec{F}_{01}|}\over{|q_0|}} \\
&= {{917.\;.\!.\!.\! \cdot 10^{-6} {~\rm N}}\over{1 \cdot 10^{-9} ~\rm C }} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm N}\over{\rm C}} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm VAs/m}\over{\rm As}} \\
&= 917.\;.\!.\!.\! \cdot 10^{3}{{\rm V}\over{\rm m}} \\
\end{align*}