Lightning Strike: Examining Electric Field, Capacitance & Power

[bkl4life]

Homework Statement

Lighting strike: Storm clouds build up large negative charges that dwell in a charge center. Suppose a storm cloud has -25 C in a charge center located 10 km above the ground. The negative charge center attracts an equal amount of positive charge that is spread on the ground below the cloud. The charge center and the ground function as a charged capacitor, with a potential difference of 4 x 10^8 V. The large electric field between these two electrodes may ionize the air leading to a conducting path between the cloud and the ground.

a) What is the magnitude of the electric field between the charge center and the ground?
b) What is the capacitance of the charge center ground system?
c) If the cloud transfers all of its charge to the ground via several rapid lighting flashes lasting a total of 1 s what is the average power.

Homework Equations

The Attempt at a Solution

I started off by using F=k(q'*q')/r^2

-25=8.99*10^9*(4 *10^8)(q')/(10,000)^2
q'=-3.29*10^-10.

Am I even on the right track?

 

[Redbelly98]

Uh, no, not really.

You're plugging in 4 *10^8 for the charge, but that number refers to the potential, not the charge.
You're plugging in -25 for the force, but that number refers to the charge, not the force.

Hint for (a): the units of electric field are Volts per meter. You are given a voltage, and a distance in meters.

(b): there should be an equation defining capacitance in your textbook. It involves voltage and charge, which are given in the problem statement.

Get those, and then we'll work on (c).

 

[nlightner]

I appreciate your attempt at applying the fundamental equation for electric force. However, in this situation, we are dealing with a charged capacitor, not just a single point charge. Therefore, we need to use different equations to calculate the electric field, capacitance, and power.

a) To calculate the electric field between the charge center and the ground, we can use the equation E = V/d, where E is the electric field, V is the potential difference, and d is the distance between the two electrodes. Plugging in the given values, we get:

E = (4 x 10^8 V) / (10 km) = 4 x 10^4 V/m

b) The capacitance of a charged capacitor is given by the equation C = Q/V, where C is the capacitance, Q is the charge, and V is the potential difference. Plugging in the values given, we get:

C = (-25 C) / (4 x 10^8 V) = -6.25 x 10^-8 F

Note that the negative sign indicates that the charge on the cloud is opposite in sign to the charge on the ground.

c) To calculate the average power, we can use the equation P = IV, where P is the power, I is the current, and V is the potential difference. In this case, the current is equal to the charge transferred divided by the time, so we have:

P = (Q/t) * V

Since the cloud transfers all of its charge (-25 C) in a total time of 1 s, the average power would be:

P = (-25 C / 1 s) * (4 x 10^8 V) = -1 x 10^10 W

Again, the negative sign indicates that the power is being transferred from the cloud to the ground. This is a very large amount of power, which is why lightning strikes can be so destructive.

In conclusion, to fully understand the electric field, capacitance, and power involved in a lightning strike, we need to consider the charged capacitor formed by the storm cloud and the ground. By applying the appropriate equations, we can calculate the magnitude of the electric field, the capacitance, and the average power involved in this phenomenon.