\(Q = \frac{{10}}{3} \times {10^{ - 9}}\ C\)
\(==>\) \(a = 8\,cm\, = \,\,8 \times {10^{ - 2}}\ m\)
\(V = 5 \times 9 \times {10^9} \times \frac{{\frac{{10}}{3} \times {{10}^{ - 9}}}}{{\frac{{8 \times {{10}^{ - 2}}}}{{\sqrt 2 }}}}\)\( = 1500\sqrt 2 \,volt\)