, the answer is 2020 kV
ExplanationElectric potential of a point charge is
[tex]\begin{gathered} V=\frac{KQ}{d} \\ where\text{ kis a constant k=9*10}^9\text{ N*}\frac{m^2}{C^2} \\ Q\text{ is charge , d is the distance} \end{gathered}[/tex]
so
Step 1
Diagram:
Step 2
distances:fromthe center to each charge:
:
[tex]\begin{gathered} d=\sqrt{0.10^2+0.10^2}=\sqrt{0.02}\text{ m=0.1414m} \\ d=0.1414\text{ m} \end{gathered}[/tex]
Step 3
electric potential due to each charge.
a) Q1
[tex]\begin{gathered} V=\frac{KQ}{d} \\ V=\text{9*10}^9\text{ N*}\frac{m^2}{C^2}*\frac{10*10^{-6}C}{0.1414\text{ m}} \\ V=636.49\text{ kV} \\ \end{gathered}[/tex]
b)Q2
[tex]\begin{gathered} V=\frac{KQ}{d} \\ V=\text{9*10}^9\text{ N*}\frac{m^2}{C^2}*\frac{-5*10^{-6}C}{0.1414\text{ m}} \\ V=-318.24\text{ KV} \\ \end{gathered}[/tex]
c)Q3
[tex]\begin{gathered} V=\frac{KQ}{d} \\ V=\text{9*10}^9\text{ N*}\frac{m^2}{C^2}*\frac{-15*10^{-6}C}{0.1414\text{ m}} \\ V=-954.73\text{kV} \\ \end{gathered}[/tex]
d)Q4
[tex]\begin{gathered} V=\frac{KQ}{d} \\ V=\text{9*10}^9\text{ N*}\frac{m^2}{C^2}*\frac{20*10^{-6}C}{0.1414\text{ m}} \\ V=-954.73\text{kV} \\ V=1272.98\text{ kV} \end{gathered}[/tex]
so, the resultant electric potential are:
:
So
[tex]\begin{gathered} v=\sqrt{1309.2^2+1591.22^2} \\ v=2060\text{ Kv} \end{gathered}[/tex]
therefore, the answer is 2020 kV
I hope this helps you
