Electric potential energy in a lightnight bolt

[estow217]

1. During a particular thunderstorm, the electric potential difference between the clouds and the ground is Vcloud-Vground = 10^9 Volts with the cloud being at the higher potential. Now, there is a lightning flash in which a charge of -25C is transferred from the groud to the cloud.

2. a) how much work is done on the charge by the electric force?

b) if the work done by the electric force were instead used to accelerate a 1100kg automobile from rest, what would be its final speed?

c) if the work done were converted to heat, how many kiligrams of water at room temperature could be vaporized?


3. I know that the energy required to vaporize a certain mas of water is Q = cmdeltaT + mL. The specific heat capacity of water at room temparature is apporximately c = 4186J/(kgxC). The latent heat capacity of water is L = 22.6 x 10^5J/kg. The temperature change is deltaT(100-20)C.
I'm really not sure how to start on part a but will I need to use the work energy theorem for part b?

 

[jbsteele]

a) The work done on the charge by the electric force is W = qVcloud-Vground = (-25C)(10^9V) = -2.5x10^11 J. b) The final speed of the automobile after the work done by the electric force is applied can be calculated using the work-energy theorem: W = KEf = 1/2mv^2f, where m is the mass of the automobile. Solving for vf, we have vf = sqrt(2W/m) = sqrt(2(-2.5x10^11J)/(1100kg)) = 587 m/s.c) To calculate the number of kilograms of water that could be vaporized with the work done by the electric force, we can use the equation Q = cmdeltaT + mL, where c is the specific heat capacity of water, deltaT is the temperature change (100-20)C, and L is the latent heat capacity of water. Thus, Q = (4186J/(kgxC))(80C) + (22.6 x 10^5J/kg) = 2.3x10^7J/kg. Therefore, the number of kilograms of water that could be vaporized is 2.5x10^11J / 2.3x10^7J/kg = 10.9 kg.

 

[kerimek]

I would like to provide a response to the content by first explaining the concept of electric potential energy in a lightning bolt. Electric potential energy is the energy that a charged object possesses due to its position in an electric field. In the case of a lightning bolt, this energy is stored in the form of electric potential difference between the clouds and the ground.

Now, let's consider the specific scenario given in the content. The electric potential difference between the cloud and the ground is 10^9 Volts, with the cloud being at the higher potential. This means that the cloud has a greater amount of electric potential energy compared to the ground. When a lightning flash occurs, a charge of -25C is transferred from the ground to the cloud. This transfer of charge results in a release of electric potential energy in the form of a lightning bolt.

To answer the questions posed in the content, we first need to understand the relationship between electric potential energy, electric force, and work. The electric force is the force exerted on a charged object by an electric field, and work is the energy transferred when a force is applied over a certain distance. In this case, the electric force is responsible for the transfer of charge and the release of electric potential energy in the form of a lightning bolt.

a) To calculate the work done on the charge by the electric force, we can use the formula W = qΔV, where q is the charge and ΔV is the change in electric potential. In this case, the work done would be -25C x (10^9V - 0V) = -2.5 x 10^10 Joules.

b) To find the final speed of the 1100kg automobile, we can use the work-energy theorem, which states that the work done on an object is equal to the change in kinetic energy of the object. In this case, the work done by the electric force (-2.5 x 10^10 Joules) would be equal to the change in kinetic energy of the automobile, which can be calculated using the formula KE = 1/2mv^2. Solving for v, we get a final speed of approximately 1.9 x 10^5 meters per second.

c) To determine the amount of water that could be vaporized by the work done, we can use the formula Q = cmdeltaT + mL, where Q is the energy