Chapter 17, Problem 68GP

To determine

The difference in the speed of the two balls when they hit the ground.

Answer to Problem 68GP

  $\text{0}\text{.12}\text{m/s}$

Explanation of Solution

Given:

Charges, ${\text{q}}_{\text{1}}\text{=}\text{650}\text{μC}\text{=}\text{6}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{C}$ , ${\text{q}}_{\text{2}}\text{=-}\text{650}\text{μC}\text{=-}\text{6}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{C}$

Mass, $\text{m}\text{=}\text{0}\text{.540}\text{kg}$

Height, $\text{h}\text{=}\text{2}\text{.00}\text{m}$

Formula used:

  $\text{E}\text{=}\text{mgh}$

Where,

  $\text{g}$ is the acceleration due to gravity.

h is the height

  $\text{m}$ is mass of the body

The potential energy is given by

  $\text{PE = qV}$

Where,

  $\text{q}$ is the charge of an electron.

  $\text{V}$ is the voltage applied.

The kinetic energy is given by $\text{KE=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}$

Where,

  $\text{m}$ is the mass of an electron.

  $\text{v}$ is the velocity of the electron.

Calculation:

Since the electric field points downward the surface of the Earth is a lower potential than points above the surface.

Given that the surface of the Earth has zero volts and height of $\text{2}\text{.00}\text{m}$ has a potential of $\text{300}\text{V}$ .

Also, the surface of the Earth has zero location for the gravitational potential energy.

So, the conservation of energy relating the charge spheres at $\text{2}\text{.00}\text{m}$ and at ground level.

  $\begin{array}{l}{\text{E}}_{\text{initial }}\text{=}{\text{E}}_{\text{final}}\\ \Rightarrow \text{mgh}\text{+}\text{qV}\text{=}\frac{\text{1}}{\text{2}}{\text{mv}}^{\text{2}}\\ \Rightarrow \text{v}\text{=}\sqrt{\text{2}\left(\text{gh+}\frac{\text{qV}}{\text{m}}\right)}\end{array}$

Speed for positive charge:

  ${\text{v}}_{\text{+}}\text{=}\sqrt{\text{2}\left(\left(\text{9}\text{.80}{\text{m/s}}^{\text{2}}\right)\left(\text{2}\text{.00}\text{m}\right)\text{+}\frac{\left(\text{6}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{C}\right)\left(\text{300}\text{V}\right)}{\left(\text{0}\text{.540}\text{kg}\right)}\right)}\text{=}\text{6}\text{.3184}\text{m/s}$

Speed for negative charge:

  ${\text{v}}_{\text{-}}\text{=}\sqrt{\text{2}\left(\left(\text{9}\text{.80}{\text{m/s}}^{\text{2}}\right)\left(\text{2}\text{.00}\text{m}\right)\text{+}\frac{\left(\text{-6}\text{.5}\text{×}{\text{10}}^{\text{-4}}\text{C}\right)\left(\text{300}\text{V}\right)}{\left(\text{0}\text{.540}\text{kg}\right)}\right)}\text{=}\text{6}\text{.2030}\text{m/s}$

Now, find the difference between both velocity:

  ${\text{v}}_{\text{+}}\text{-}{\text{v}}_{\text{-}}\text{=}\text{6}\text{.3184}\text{m/s}\text{- 6}\text{.2030}\text{m/s}\text{=}\text{0}\text{.12}\text{m/s}$

Conclusion:

The required value is $\text{0}\text{.12}\text{m/s}$ .