From the diagram we can notice that the motion is along x-axis.

\(\displaystyle{x}_{{i}}={25}{c}{m},{x}_{{f}}={10}{c}{m}\)

Consider \(\displaystyle{x}_{{i}}={0}\) and the positve x-axis toward the 3nC charge, so \(\displaystyle{x}_{{f}}={15}{c}{m}\).

\(\displaystyle{x}_{{f}}-{x}_{{i}}={15}{c}{m}\)

F at this point equal to

\(\displaystyle\sum{F}={F}_{{{b}{y}{3}{n}{C}}}+{F}_{{{b}{y}{2}{n}{C}}}\)

\(\displaystyle{F}_{{{b}{y}{3}{n}{C}}}={k}{\frac{{{3}\times{10}^{{-{6}}}-{1.6}\times{10}^{{-{19}}}}}{{{10}}}}\times{10}^{{-{2}}}\)

\(\displaystyle{F}_{{{b}{y}{2}{n}{C}}}={k}{\left({2}\times{10}^{{-{6}}}-{1.6}\times{10}^{{-{19}}}\right\rbrace}{\left\lbrace{40}\right\rbrace}\times{10}^{{-{2}}}\)

Caculate both forces and then ,

\(\displaystyle{a}{x}={\frac{{\sum{F}}}{{{m}_{{{e}le{c}{t}{r}{o}{n}}}}}}\)

when u get the acceleration u can get the velocity usingconstant acceleration motion equations ,

Use :

\(\displaystyle{v}_{{x}}=\sqrt{{{2}{a}{x}{\left({x}{f}-\xi\right)}}}\)

u know that the initial velocity is zero as long we assumedthe motion started from the origin

and \((xf-xi) = 15cm\)