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Chapter 17: Problem 66

Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?

Short Answer

Expert verified
Answer: Approximately \(300.3 \mathrm{m}\).

Step by step solution

01

Understand the relation between electric field, voltage difference, and distance

An electric field (E) can create a potential difference (V) which is the product of the electric field and the distance (d) between the points. Mathematically, it is given by: \(V = E \times d\) where V is the potential difference, E is the electric field strength, and d is the distance between the points
02

Use given values for electric field and voltage difference

The electric field strength at which the breakdown limit of damp air is reached (E) is given as \(3.33 \times 10^{5} \mathrm{V/m}\). The potential difference between the bottom of the thundercloud and Earth's surface is given as \(1.00 \times 10^{8} \mathrm{V}\).
03

Rearrange the equation to find distance (d)

To find the maximum possible height (d) at which a thundercloud can exist, we can rearrange the equation from step 1 to solve for d, given as: \(d = \frac{V}{E}\) Substitute the values of V and E into the equation: \(d = \frac{1.00 \times 10^{8} \mathrm{V}}{3.33 \times 10^{5} \mathrm{V/m}}\)
04

Calculate the maximum height (d)

Divide the given numbers to find the maximum height: \(d = \frac{1.00 \times 10^{8}}{3.33 \times 10^{5}}\) \(d = 300.3 \mathrm{m}\) So, the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential \(1.00 \times 10^{8} \mathrm{V}\) below Earth's surface potential, to have a lightning strike is approximately \(300.3 \mathrm{m}\).

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Most popular questions from this chapter

The potential difference across a cell membrane from outside to inside is initially at \(-90 \mathrm{mV}\) (when in its resting phase). When a stimulus is applied, Na" ions are allowed to move into the cell such that the potential changes to \(+20 \mathrm{mV}\) for a short amount of time. (a) If the membrane capacitance per unit area is $1 \mu \mathrm{F} / \mathrm{cm}^{2},\( how much charge moves through a membrane of area \)0.05 \mathrm{cm}^{2} ?\( (b) The charge on \)\mathrm{Na}^{+}\( is \)+e$ How many ions move through the membrane?
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It has only been fairly recently that 1.0 -F capacitors have been readily available. A typical 1.0 -F capacitor can withstand up to \(5.00 \mathrm{V}\). To get an idea why it isn't easy to make a 1.0 -F capacitor, imagine making a \(1.0-\mathrm{F}\) parallel plate capacitor using titanium dioxide \((\kappa=90.0\) breakdown strength \(4.00 \mathrm{kV} / \mathrm{mm}\) ) as the dielectric. (a) Find the minimum thickness of the titanium dioxide such that the capacitor can withstand \(5.00 \mathrm{V}\). (b) Find the area of the plates so that the capacitance is \(1.0 \mathrm{F}\)
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