Final common potential

\(v=\frac{220 \times 5+0 \times 2.5}{5+2.5}=220 \times \frac{2}{3}\)

\(u _{ f }=\frac{1}{2}(5+2.5) \times 10^{-6}\left(220 \times \frac{2}{3}\right)^{2}\)

\(\Delta u = u _{ f }- u _{ i }\)

\(\Delta u =-403.33 \times 10^{-4}\)

\(\Rightarrow-403.33 \times 10^{-4}=\frac{ X }{100}\)

\(X=-4.03\)

or magnitude or value of \(X\) is approximate \(4\)