Question

Suppose that in a lightning flash the potential difference between a cloud and the ground is $\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{}\mathit{V}$and the quantity of charge transferred is 30 C. (a) What is the change in energy of that transferred charge? (b) If all the energy released could be used to accelerate a 1000 kgcar from rest, what would be its final speed?

Short answer

1. The change in energy is $∆U=3.0\times {10}^{10}J$.
2. The final speed of the car is $v=7.7\times {10}^{3}m/s$.

Step-by-step solution

Given data:

• The potential difference between a cloud and the ground is $∆V=1.0\times {10}^{9}V$.
• The quantity of charge transferred is q = 30 C.
• The mass of the car is m = 1000 kg.

Understanding the concept:

Using the equation of potential energy U = qV, you can find change in potential energy, you can find change in energy of the transferred charge.

Using the conservation of energy, you can find the car's final speed.

(a) Calculate the change in energy of that transferred charge:

The change in energy of the transferred charge is.

$\begin{array}{rcl}∆U& =& q∆V\\ & =& \left(30C\right)\left(1.0\times {10}^{9}V\right)\\ & =& 3.0\times {10}^{10}J\end{array}$

Hence, the change in energy is $\begin{array}{rcl}& & 3.0\times {10}^{10}J\end{array}$.

(b) Calculate the final speed if all the energy released could be used to accelerate a 1000 kg car from rest:

If all this energy is used to accelerate a 1000 kg car from rest, then the energy will be,

$∆U=K=\frac{1}{2}m{v}^2$

And you can find the car’s final speed to be

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

Substitute known values in the above equation.

role="math" localid="1662533780803" $\begin{array}{rcl}V& =& \sqrt{\frac{2\left(3.0\times {10}^{10}J\right)}{1000kg}}\\ & =& \sqrt{60\times {10}^6{m}^2/s^2}\\ & =& 7.7\times {10}^{3}m/s\end{array}$

Hence, the final speed of the car is $\begin{array}{rcl}& & 7.7\times {10}^{3}m/s\end{array}$.