Chapter 2: Problem 30
For which values of \(U\) is the magnitude of \(\Phi_{\text {cir }}\) independent of the radius \(r\) ?
Short Answer
Expert verified
Answer: The magnitude of \(\Phi_{\text{cir}}\) is independent of the radius r for all constant values of U.
Step by step solution
01
Identify the expression for \(\Phi_{\text {cir }}\)
We need the expression for the magnitude of \(\Phi_{\text {cir }}\). It is given by:
\(\Phi_{\text {cir }} = \int_{0}^{2\pi} U(\vec{r})\ \vec{t}\ ds \)
Here, \(U(\vec{r})\) is the radial electric field, \(\vec{t}\) is the unit tangent vector, and \(ds\) is the infinitesimal length along the path.
02
Write the expression for the radial electric field
We are given that the radial electric field is \(U\). Let's write the electric field as a function of r:
\(U(\vec{r}) = U(r)\)
03
Calculate the unit tangent vector
As we are considering a circular path, the unit tangent vector \(\vec{t}\) is given by:
\(\vec{t} = \frac{d\vec{r}}{ds} = \frac{1}{r}\frac{d\vec{r}}{d\theta}\)
Where \(\theta\) is the angle measured from the positive x-axis.
04
Compute \(ds\) for the circular path
For a circular path, the infinitesimal length \(ds\) is given by the arc length formula:
\(ds = r\ d\theta\)
05
Calculate the magnitude of \(\Phi_{\text {cir }}\)
Now we substitute the expressions for \(U(r)\), \(\vec{t}\), and \(ds\) into the integral for \(\Phi_{\text {cir }}\):
\(\Phi_{\text {cir }} = \int_{0}^{2\pi} U(r)\frac{1}{r}\frac{d\vec{r}}{d\theta}\ r\ d\theta\)
This simplifies to:
\(\Phi_{\text {cir }} = \int_{0}^{2\pi} U(r)\frac{d\vec{r}}{d\theta}\ d\theta\)
Since \(\frac{d\vec{r}}{d\theta}\) is independent of r, we can pull the \(U(r)\) term out of the integral:
\(\Phi_{\text {cir }} = U(r)\int_{0}^{2\pi} \frac{d\vec{r}}{d\theta}\ d\theta\)
This means the magnitude of \(\Phi_{\text {cir}}\) is independent of r when \(U(r)\) is a constant value.
06
Find the values of U for which \(\Phi_{\text {cir }}\) is independent of r
Since the magnitude of \(\Phi_{\text{cir}}\) is independent of r when \(U(r)\) is a constant value, the answer is:
For all values of U that are constants, the magnitude of \(\Phi_{\text{cir}}\) is independent of the radius r.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Radial Electric Field
Imagine standing in a vast, open field with a single strong magnet at the center. Similar to how a magnet creates a field around it, electric charges create an electric field in the space around them. A radial electric field is a special type of electric field that emanates outward from a point charge, just like the sun's rays spreading out in all directions.
The strength of a radial electric field emanating from a point charge decreases with distance from the charge, akin to how the warmth of the sun feels less intense the further away you are. Mathematically, we often express this field as a function of the distance, denoted by the variable r. In problems like the one under consideration, if the electric field's intensity U remains constant regardless of r, it implies that the influence or 'push' experienced by a charge within this field is the same no matter how far away it is from the source.
When problem-solving, this characteristic can significantly simplify calculations. If U(r) does not change with r, it means that the integral calculations over a path can be simplified as the variable field component does not need to be considered.
The strength of a radial electric field emanating from a point charge decreases with distance from the charge, akin to how the warmth of the sun feels less intense the further away you are. Mathematically, we often express this field as a function of the distance, denoted by the variable r. In problems like the one under consideration, if the electric field's intensity U remains constant regardless of r, it implies that the influence or 'push' experienced by a charge within this field is the same no matter how far away it is from the source.
When problem-solving, this characteristic can significantly simplify calculations. If U(r) does not change with r, it means that the integral calculations over a path can be simplified as the variable field component does not need to be considered.
Unit Tangent Vector
Picture yourself on a merry-go-round, holding tight to a horse as it circles around. Your path on the merry-go-round traces a circle, similar to the circular path we encounter in this problem. The direction you're facing at any point on that ride can be thought of as a unit tangent vector, designated as vec{t} in our calculations.
Mathematically, the unit tangent vector is crucial as it tells us the direction of a path at a particular point, offering directionality to the movement. In a circular path, this vector is always perpendicular to the radius and points in the direction of motion along the circle's edge. Because it's a 'unit' vector, its length is normalized to one, meaning it gives us pure direction without concern for how 'strong' or 'fast' the path is being traced.
Mathematically, the unit tangent vector is crucial as it tells us the direction of a path at a particular point, offering directionality to the movement. In a circular path, this vector is always perpendicular to the radius and points in the direction of motion along the circle's edge. Because it's a 'unit' vector, its length is normalized to one, meaning it gives us pure direction without concern for how 'strong' or 'fast' the path is being traced.
Importance in Calculations
When computing integrals of vector fields along a path, the unit tangent vector aligns the field to the direction of the path, ensuring that only the component of the field that's parallel to the path is measured. This alignment is what gives the unit tangent vector its significance in line integrals.Line Integral of Vector Field
Now combine your understanding of both a radial electric field and the unit tangent vector as we explore the concept of the line integral of a vector field. It's like adding up all the tiny pushes and pulls you feel as you walk through the electric field around a charged object, following a specific path.
The formal definition of a line integral may sound complex, but in essence, it's just a method to tally up these effects along a curve or path. For our circular path, the line integral calculates how the radial electric field's 'push' interacts with the circular motion we're considering, quantified over the entire loop. It’s like counting each step's contribution while walking along a path.
The formal definition of a line integral may sound complex, but in essence, it's just a method to tally up these effects along a curve or path. For our circular path, the line integral calculates how the radial electric field's 'push' interacts with the circular motion we're considering, quantified over the entire loop. It’s like counting each step's contribution while walking along a path.