Question

An arrangement of source charges produces the electric potential V = 5000x2 along the x-axis, where V is in volts and x is in meters. What is the maximum speed of a 1.0 g, 10 nC charged particle that moves in this potential with turning points at 8.0 cm?

Short answer

The value of maximum speed of a 1.0 g, 10 n C charged particle 2.53 cm/s

Step-by-step solution

The theory of the maximum speed of a  charged particle 

The potential energy as potential function is as follows;

$U=qV$

Now the expression of potential is;

$U\left(x\right)=5000q{x}^2$

There is no potential energy, all energy here is potential. Thus after application law of energy conversion;

$\begin{array}{rcl}U& =& \frac{m{v}^2}{2}\\ v& =& \sqrt{\frac{2U}{m}}\end{array}$

Formulation of kinetic energy

The value of maximum speed of a 1.0 g, 10 n C charged particle

$v\left(x\right)=\sqrt{\frac{{10}^{4}q{x}^2}{m}}$

After substituting all the values speed of particle is;

$\begin{array}{rcl}v& =& \sqrt{\frac{{10}^4\times {10}^{-8}\times 0.{08}^2}{0.001}}\\ & =& 0.025m/s\end{array}$