91Ó°ÊÓ

The electric potential energy and electric Potential difference the related through given equation;

U= qV

Here in this Question given Mass particle is;

$m=1.0g=1.0×{10}^{-3}Kg$

And, the charge of particle is;

$q=10.0nC=1.0×{10}^{-8}C$

Due the arrangement of the sources electric charges which are produced are;

$V=5000x^2$

Here in this Equation V is in Volt and X is in meters

As charged particle moves at turning point at 8cm

Which denotes that;

$X_{max}=Â\pm 8cm$

At this Point Partcle will have only Potential energy, there will be no Kinetic energy.

Thus the Max Potential energy is;

$$\begin{gathered} U_{max}=qV \\ ApplingtheGivenDatawithinthisEquation; \\ {U}_{max}=(1.0×{10}^{-8}C)\left(5000\right)\left({(8×{10}^{-2}m)}^2\right) \\ {U}_{max}=3.20×{10}^{-7}J \end{gathered}$$

At the time hence the particle comes back to the origin x=0, the potential energy converted into kinetic energy

$$\begin{gathered} Hence;\frac{1}{2}m{v^2}_{max}=\sqrt{\frac{2U_{max}}{m}} \\ Replacing,1.0×{10}^{-3}Kginplaceofmassand3.20×{10}^{-7}Jinplaceof{U}_{max} \\ {v}_{max}=\sqrt{\frac{2(3.20×{10}^{-7}J)}{1.0×{10}^{-3}kg}} \\ ThustheMaximumSpeedofParticleis{v}_{max}=2.53cm/s \end{gathered}$$