Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0 \times 10^{9}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 kg car from rest, what would be its final speed?

# Solution:

### a) To find the change of energy of the transferred charge, we multiply the potential difference by the quantity of charge transferred.

### \triangle U = q \triangle V = (30 C)(1.0 \times 10^{9} V) = 3.0 \times 10^{10} J

### b) To figure out the velocity of the truck, we must use the kinetic energy equation. We know that all the energy released is used to accelerate the truck, therefore, \triangle U=K=\frac{1}{2}mv^{2}.

### v = \sqrt{\frac{2K}{m}}=\sqrt{\frac{2\triangle U}{m}}

### = \sqrt{\frac{2(3.0 \times 10^{10}J)}{1000kg}}=7.7 \times 10^{3}\frac{m}{s}

What about if energy released could be used to melt ice, how much ice would it melt? Take latent heat of fusion for ice to be 333.5KJ/kg

Correct me if I am wrong, but,

The quantity of heat Q supplied or given out during a change of state is given by:

Q = mL

(where m is the mass in kilograms and L is the specific latent heat.)

So if we go by that equation, the lighting strike produces 3.0×10^(10) J.

Therefore, m=Q/L

m=(3.0×10^(10)J)/(333.5×10^(3)J/kg)

m = 89,995 kg of ice melted.

It was helpful. Thank you

Glad it helped! Good luck with your studies 🙂

formula of energy is 1/2 QV………..and u used QV…??????.

Not sure where you got your equation from. The one used in the solution is the correct one.

After using this solution to answer a homework question of my own, I tried the suggested 1/2QC suggested in the comment section and got the question correct. I would suggest updating your answer to reflect the correct equation.

Potential energy of 2 point charges is u=qv and electrostatic potential energy is u=1/2qv. You have to know which equation to use when it’s appropriate. It’s important to understand why these work at certain situations rather than blindly following a single equation to get an answer. The answer in this solution is the correct one.