# Capacitors

## Practice

### practice problem 1

#### solution

Answer it.

### practice problem 2

#### solution

Answer it.

### practice problem 3

*tropos*) meaning turn, a turnaround, or

*a change*. The troposphere is that part of the atmosphere where all the day to day

*changes*we call weather occur. One of the side effects of thunderstorms is the separation of charge in the troposphere — positive at the top, negative at the bottom. The atmosphere behaves like a circuit with properties like voltage, current, resistance, and capacitance. Given the title of this section, we'll be concentrating on the capacitance. To start this problem, you need to review some old physics related to fluids. To finish it, you need to apply some new physics related to electrostatics.

- Determine the effective thickness of the troposphere — a.k.a. its scale height. What height would the atmosphere have if it was one uniform density from bottom to top? Hint: Use surface pressure to determine the mass of the atmosphere, assume the density on the surface is the density everywhere, and determine the thickness of this ideal atmosphere. Additional hint: Figure this out using symbols and algebra until you have one equation with thickness as the subject before you start playing with numbers.
- Determine the total capacitance of the troposphere.

#### solution

This is a review of concepts presented in an earlier part of this book. Since pressure is defined as force per area, force is equal to pressure times area.

*F*=*PA*⇐ *P*=*F**A*The force in this case is the weight of the Earth's atmosphere (

*mg*).*mg*=*PA*Solve for mass…

*m*=*PA**g*…and then work on something else.

Since density is defined as the ratio of mass to volume, mass is the product of density and volume. (Remember, the symbol for density is the Greek letter rho not the Latin letter p.)

*m*= ρ*V*⇐ ρ = *m**V*Treat the atmosphere as a very thin shell on top of the surface of the Earth. Its volume would then be the surface area of the Earth (

*A*) times the thickness of the shell (*h*).*m*= ρ*Ah*Set the two mass equations equal to one another.

ρ *Ah*=*PA**g*Solve for thickness and note how nicely area cancels out.

*h*=*P*ρ *g*We're ready for numbers…

*h*=(101,325 Pa) (1.21 kg/m ^{3})(9.8 m/s^{2})…and an answer.

*h*= 8.5 kmCompare this to the radius of the whole Earth.

8.5 ^{}km≳ 0.1% 6371 km Don't think about it as "How

*thick*is the atmosphere" but rather "How*thin*is the atmosphere". A standard pack of copier or printer paper has 500 sheets. Unwrap two of them and stack one on top of the other. Remove one sheet of paper. That's the thickness of the Earth's atmosphere.The troposphere is a spherical capacitor — sort of.

*C*=4πκε _{0}(1/ *r*_{1}) − (1/*r*_{2})The reason I say sort of is because the difference between the inner radius (

*r*_{1}) and the outer radius (*r*_{2}) is almost zero. Maybe if the equation above were written slightly differfently. With a little bit of algebra, the weird compound fraction above can be written as the slightly less weird fraction below.*C*= 4πκε_{0}*r*_{1}*r*_{2}*r*_{2}−*r*_{1}And now for some algebra they don't teach you in algebra class. Since the two radii are nearly the same, their product is essentially the square of one number (let's call it

*r*^{2}) and their difference is something we've been describing as a thickness (*h*). That gives us an equation that's now only approximately true when*h*≪*r*(*h*is much smaller than*r*).*C*≈ 4πκε_{0}*r*^{2}/*h*We're ready for numbers…

*C*= 4π(8.85 × 10^{−12}C^{2}/N m^{2})(6,371,000 m)^{2}/(8,500 m)…and an answer.

*C*= 0.53 F

### practice problem 4

#### solution

Answer it.