From the figure, we have
OP = OQ = OR = OS = OT = OU
= r = 10 cm = 0.1m
And given q = 5µC = 5 × 10-6 C
∴ Potential at O due to all the charges
\(v=6\times \frac{1}{4\pi \varepsilon_0}\frac{q}{r}\) = \(\frac{6\times 9\times10^9\times5\times10^{-6}}{0.1}\)
= 2.7 x 106 volt