Chapter 25: Q. 45 (page 711)

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?

Expert verified

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

Step by step solution

The potential energy as potential function is as follows;

$U = q V$

Now the expression of potential is;

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

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

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

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

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

After substituting all the values speed of particle is;

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

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