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Chapter 24: Problem 47

Fair Weather During fair weather, an electric field of about 100 \(\mathrm{N} / \mathrm{C}\) points vertically downward into Earth's atmosphere. Assuming that this field arises from charge distributed in a spherically symmetric manner over the surface of Earth, determine the net charge of Earth and its atmosphere if the radius of Earth and its atmosphere is \(6.37 \times 10^{6} \mathrm{~m}\).

Short Answer

Expert verified
\(4.5 \times 10^5 \ \text{C}\)

Step by step solution

01

Understand the Given Data

The given electric field is 100 \(\text{N/C}\) pointing downward. The radius of Earth is \(6.37 \times 10^6 \ \text{m}\). We need to find the net charge of Earth and its atmosphere.
02

Recall the Electric Field Formula

Recall the formula for the electric field due to a spherical charge distribution: \[E = \dfrac{1}{4 \pi \epsilon_0} \dfrac{Q}{r^2} \] where \(E\) is the electric field, \(Q\) is the net charge, \(r\) is the radius, and \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 = 8.85 \times 10^{-12} \ \text{F/m} \)).
03

Rearrange the Formula

Rearrange the formula to solve for the net charge \(Q\): \[Q = E \cdot 4 \pi \epsilon_0 \cdot r^2 \]
04

Input the Given Values

Substitute the given values into the rearranged formula: \[ Q = 100 \cdot 4 \pi \cdot 8.85 \times 10^{-12} \cd \ (6.37 \times 10^6)^2 \]
05

Perform the Calculations

Perform the calculation step-by-step: \[(6.37 \times 10^6)^2 = 4.05869 \times 10^{13} \] \[ 100 \cdot 4 \pi \cdot 8.85 \times 10^{-12} \cdot 4.05869 \times 10^{13} = 4.5 \times 10^5 \ \mathrm{C} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electrostatics
Electrostatics is the study of electric charges at rest. It's all about understanding how electric charges interact when they're not in motion. In electrostatics, we explore the laws governing electric forces and fields created by stationary charges.
When you have charges accumulated in one place, their interactions can produce electric fields. An electric field is a region around a charged particle where other charges experience a force.
The strength and direction of this electric field at any point can be determined using the Coulomb's law and electric field equations.
  • The electric field due to a point charge is given by: \(E = k \frac{Q}{r^2}\), where \(E\) is the electric field, \(Q\) is the charge, \(r\) is the distance from the charge, and \(k\) is Coulomb's constant.
  • In our exercise, Earth's electric field points downward, meaning it's affecting charges above Earth's surface.
Net Charge
The net charge of a system is the sum of all electric charges within that system. Charges can be positive or negative. When calculating the net charge, you take into account both.
To find the net charge of Earth and its atmosphere, we use the formula for the electric field created by a spherical charge distribution. From electrostatics, we know the electric field \(E\) at the surface of a sphere of radius \(r\) is given by:
\[E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}\]
  • Here, \(Q\) is the net charge causing the field, and \(\epsilon_0\) is the permittivity of free space.
  • By rearranging this equation, we find the net charge as: \[Q = E \cdot 4 \pi \epsilon_0 \cdot r^2\]
This step ensures we can calculate \(Q\) with the given electric field \(E\) and radius \(r\).
Spherical Charge Distribution
A spherical charge distribution assumes that the charge is uniformly spread over a spherical surface. This assumption simplifies the mathematics for problems involving electric fields.
In our exercise, we deal with such a spherical distribution over Earth鈥檚 surface. The electric field due to a uniformly charged sphere can be treated as if all the charge were concentrated at the center:
\[E = \frac{Q}{4 \pi \epsilon_0 \cdot r^2}\]
  • This formula allows us to calculate the electric field at distance \(r\) from the sphere鈥檚 center.
  • We assume the electric field given (100 N/C) is due to this symmetrical charge distribution.
By using the Earth's radius and given electric field, we derive the net charge. This scenario is common in problems involving planetary electric fields and helps understand similar astrophysical contexts.

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Most popular questions from this chapter

Infinite Line of Charge An infinite line of charge produces a field magnitude of \(4.5 \times 10^{4} \mathrm{~N} / \mathrm{C}\) at a distance of \(2.0 \mathrm{~m} .\) Calculate the amount of linear charge density \(|\lambda|\).

Charge Causes Flux A point charge causes an electric flux of \(-750 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}\) to pass through a spherical Gaussian surface of \(10.0 \mathrm{~cm}\) radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge?

Charge at Center of Shell A point charge \(+q\) is placed at the center of an electrically neutral, spherical conducting shell with inner radius \(a\) and outer radius \(b .\) What charge appears on (a) the inner surface of the shell and (b) the outer surface? What is the net electric field at a distance \(r\) from the center of the shell if (c) \(rr>a\), and (e) \(r>b\) ? Sketch field lines for those three regions. For \(r>b\), what is the net electric field due to (f) the central point charge plus the inner surface charge and (g) the outer surface charge? A point charge \(-q\) is now placed outside the shell. Does this point charge change the charge distribution on (h) the outer surface and (i) the inner surface? Sketch the field lines now. (j) Is there an electrostatic force on the second point charge? (k) Is there a net electrostatic force on the first point charge? (1) Does this situation violate Newton's Third Law?

Thin Metal Plates Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of opposite signs. The amount of charge per unit area is given by \(|\sigma|=7.0 \times 10^{-22} \mathrm{C} / \mathrm{m}^{2}\), with the negatively charged plate on the left. What are the magnitude and direction of the electric field \(\vec{E}\) (a) to the left of the plates, (b) to the right of the plates, and (c) between the plates?

Shower When a shower is turned on in a closed bathroom, the splashing of the water on the bare tub can fill the room's air with negatively charged ions and produce an electric field in the air as great as \(1000 \mathrm{~N} / \mathrm{C}\). Consider a bathroom with dimensions of \(2.5 \mathrm{~m} \times 3.0 \mathrm{~m} \times 2.0 \mathrm{~m}\). Along the ceiling, floor, and four walls, approximate the electric field in the air as being directed perpendicular to the surface and as having a uniform magnitude of \(600 \mathrm{~N} / \mathrm{C}\). Also, treat those surfaces as forming a closed Gaussian surface around the room's air. What are (a) the volume charge density \(\rho\) and (b) the number of excess elementary charges \(e\) per cubic meter in the room's air?
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