Suppose that in a lightning flash the potential difference 8

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?


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}


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8 thoughts on “Suppose that in a lightning flash the potential difference

    • Steven

      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 = 89,995 kg of ice melted.

  • Matt Soby

    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.

    • questionsolutions Post author

      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.