Problem 56
Question
The strength \(E\) of an electric field at point \((x, y, z)\) resulting from an infinitely long charged wire lying along the \(y\) -axis is given by \(E(x, y, z)=k / \sqrt{x^{2}+y^{2}},\) where \(k\) is a positive constant. For simplicity, let \(k=1\) and find the equations of the level surfaces for \(E=10\) and \(E=100\).
Step-by-Step Solution
Verified Answer
Level surfaces equations are \(x^2 + z^2 = \frac{1}{100}\) for \(E=10\) and \(x^2 + z^2 = \frac{1}{10000}\) for \(E=100\).
1Step 1: Understand Level Surfaces
Level surfaces are surfaces where the function, in this case the strength of the electric field \(E(x, y, z)\), takes on a constant value. Here, we need to find the surfaces where \(E = 10\) and \(E = 100\).
2Step 2: Substitute Variables in Electric Field Equation
Given the function \(E(x, y, z) = \frac{1}{\sqrt{x^2 + z^2}}\) (since the wire lies along the \(y\)-axis), substitute \(k = 1\) and set the expression equal to 10 for \(E = 10\): \(\frac{1}{\sqrt{x^2 + z^2}} = 10\).
3Step 3: Solve for Level Surface Equation when E = 10
Solve the equation \(\frac{1}{\sqrt{x^2 + z^2}} = 10\) by rearranging it to \(\sqrt{x^2 + z^2} = \frac{1}{10}\), thus finding the level surface equation \(x^2 + z^2 = \frac{1}{100}\).
4Step 4: Substitute Variables for E = 100
Use the same process for \(E = 100\): Set \(\frac{1}{\sqrt{x^2 + z^2}} = 100\) and rearrange to find the equation of the level surface.
5Step 5: Solve for Level Surface Equation when E = 100
Solve the equation \(\frac{1}{\sqrt{x^2 + z^2}} = 100\) by rearranging it to \(\sqrt{x^2 + z^2} = \frac{1}{100}\), thus finding the level surface equation \(x^2 + z^2 = \frac{1}{10000}\).
Key Concepts
Level SurfacesCharged WireEquations of SurfacesElectric Field Strength
Level Surfaces
Level surfaces are important in understanding how a function behaves in space. Imagine you have a function that depends on three dimensions:
- The level surface represents all points in space where this function takes the same value.
- For example, consider the surface of water in a perfectly still swimming pool. It is at the same height everywhere, representing a level surface.
- In the context of electric fields, a level surface means any point on this surface has the same electric field strength.
Charged Wire
A charged wire, especially when imagined to be infinitely long, creates a distinctive pattern of electric fields around it. Here are its key features:
- The wire itself, charged with electricity, generates electric fields emanating outward in a circular pattern.
- For simplification, if the wire is aligned along the y-axis, the fields will spread out in the xz-plane, displaying symmetry about the wire.
- The electric field intensity decreases with the square of the distance from the wire, being stronger closer and weaker farther away.
Equations of Surfaces
Equations of surfaces help us to mathematically describe the shape and position of these level surfaces in three-dimensional space. Let's explore this further:
- The equation derived from the electric field formula helps identify surfaces where the electric field magnitude is constant.
- For instance, if the function for electric field strength is given by \( E(x, y, z) = \frac{1}{\sqrt{x^2 + z^2}} \), setting \( E = 10 \) gives us the equation \( x^2 + z^2 = \frac{1}{100} \).
- This equation represents a flat circular surface in the xz-plane at equal distance from the wire, indicating a constant electric field magnitude.
Electric Field Strength
Electric field strength, denoted as \( E \), provides an understanding of how strong the electric forces are at a given point. Here are the essentials:
- It's a measure of the force that a unit positive charge would experience at a particular location.
- The strength depends on factors like distance from the source charge and the configuration of these charges.
- Especially in the context of a linear charge like our wire, the electric field decreases with increased distance from the source, following an inverse relationship.
Other exercises in this chapter
Problem 54
For the following exercises, find an equation of the level curve of \(f\) that contains the point \(P\). $$g(x, y)=y^{2} \arctan x, P(1,2)$$
View solution Problem 55
For the following exercises, find an equation of the level curve of \(f\) that contains the point \(P\). $$g(x, y)=e^{x y}\left(x^{2}+y^{2}\right), P(1,0)$$
View solution Problem 57
A thin plate made of iron is located in the \(x y\) -plane. The temperature \(T\) in degrees Celsius at a point \(P(x, y)\) is inversely proportional to the squ
View solution Problem 60
For the following exercises, find the limit of the function. $$\lim _{(x, y) \rightarrow(1,2)} x$$
View solution