Answer:
If all these three charges are positive with a magnitude of [tex]1.8 \times 10^{-8}\; \rm C[/tex] each, the electric potential at the midpoint of segment [tex]\rm AB[/tex] would be approximately [tex]8.3 \times 10^{3}\; \rm V[/tex].
Explanation:
Convert the unit of the length of each side of this triangle to meters: [tex]10\; \rm cm = 0.10\; \rm m[/tex].
Distance between the midpoint of [tex]\rm AB[/tex] and each of the three charges:
Let [tex]k[/tex] denote Coulomb's constant ([tex]k \approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2}[/tex].)
Electric potential due to the charge at [tex]\rm A[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm A})}[/tex].
Electric potential due to the charge at [tex]\rm B[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm B})}[/tex].
Electric potential due to the charge at [tex]\rm A[/tex]: [tex]\displaystyle \frac{k\, q}{d({\rm C})}[/tex].
While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.
Hence, the electric field at the midpoint of [tex]\rm AB[/tex] due to all these three charges would be:
[tex]\begin{aligned}& \frac{k\, q}{d({\rm A})} + \frac{k\, q}{d({\rm B})} + \frac{k\, q}{d({\rm C})} \\ &= k\, \left(\frac{q}{d({\rm A})} + \frac{q}{d({\rm B})} + \frac{q}{d({\rm C})}\right) \\ &\approx 8.99 \times 10^{9}\; \rm N \cdot m^{2} \cdot C^{-2} \\ & \quad \quad \times \left(\frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{0.050\; \rm m} + \frac{1.8 \times 10^{-8} \; \rm C}{\sqrt{3} \times (0.050\; \rm m)}\right) \\ &\approx 8.3 \times 10^{3}\; \rm V\end{aligned}[/tex].
