A material with resistivity ρis formed into a cylinder of length L and outer radius router . A c

A material with resistivity ρis formed into a cylinder of...

A material with resistivity $\rho$ is formed into a cylinder of length $L$ and outer radius $r_{\text {outer }}$. A cylindrical core with radius $r_{\text {inner }}$ is removed from the axis of this cylinder and filled with a conducting material, which is attached to a wire. The outer surface of the cylinder is coated with a conducting material and attached to another wire. (a) If the second wire has potential $V$ greater than the first wire, in what direction does the local electric field point inside of the cylinder? (b) The magnitude of this electric field is $c / r,$ where $c$ is a constant and $r$ is the distance from the axis of the cylinder. Use the relationship $V=\int \vec{E} \cdot d \vec{l}$ to determine the constant $c .$ (c) What is the resistance of this device? (d) A $1.00-\mathrm{cm}$ -long hollow cylindrical resistor has an inner radius of $1.50 \mathrm{~mm}$ and an outer radius of 3.00 $\mathrm{mm}$. The material is a blend of powdered carbon and ceramic whose resistivity $\rho$ may be altered by changing the amount of carbon. If this device should have a resistance of $6.80 \mathrm{k} \Omega,$ what value of $\rho$ should be selected?

Key Concepts

Cylindrical Geometry Integration

When dealing with conductors that have cylindrical symmetry and non-uniform distributions of electric field or current density, integration in cylindrical coordinates is used. This involves integrating over radial elements, particularly when quantities like the electric field vary as a function of the radius. This method is essential for calculating quantities such as the potential difference or total resistance in objects with hollow cylindrical shapes or varying cross-sectional areas.

Resistivity and Resistance

Resistivity is an intrinsic property of a material that quantifies how strongly it opposes the flow of electric current. Resistance, on the other hand, is the measure of a particular object's opposition to current, depending on both the resistivity of the material and its geometrical dimensions. The relationship between resistivity, geometry, and resistance is vital for designing and analyzing electronic components.

Line Integrals and Potential Difference

The relationship between electric potential difference and the electric field is given by the line integral V = ?E · dl. This fundamental concept allows us to compute the potential difference between two points by integrating the electric field along any path joining them. It is critical for determining the constant in field expressions when the magnitude of the electric field is given as a function of position.

Electric Field and Potential Gradient

The electric field is defined as the negative gradient of the electric potential. This means that the electric field always points in the direction in which the electric potential decreases most rapidly. In practical terms, when one end of a conductor is at a higher potential relative to the other, the field lines point from the higher potential region to the lower potential region.

Transcript

0:00 Hi there.

00:01 So for this problem we have a material with resistivity row.

00:05 Energy is formed into a cylindrical, into a cylinder of length l.

00:10 So we have a cylinder of length l.

00:20 And it has a resistivity row.

00:25 And it has an outer radius.

00:28 When we cut the contraceptional area of this cylinder, we will have an outer radius and a outer radius.

00:34 And an inner radius.

00:36 So we call that inner radius.

00:40 I'm going to put it in here.

00:42 We call it inner and this the out radius.

00:50 I'm going to put that in here just to see it better.

00:53 So this is the out radius.

00:58 Now that is filled with a condensual material, which is attached to the wire.

01:06 There is a wire here at the center of it.

01:10 So the outer surface of the cylinder is coated with a conducting material and attached to another wire.

01:18 So the first question for this problem is, if the second wire has potential b greater than the first wire, in what direction does the local electric fuel point inside of the cylinder? so the first thing that we need to identify, is that this problem deals with or involves resistivity.

01:46 So for that, we know that the resistance is related to the resistivity by this formula, where l is the length, in this case of the cylinder, and a is the area of the creusceptional area.

02:04 So to solve this first part, part a, we know that the other conductor is at higher potential than the inner conductor.

02:14 So the electric field points inward toward the center of the apsis.

02:20 So it will be pointing like this to the center.

02:25 So that's part a of this problem.

02:28 We need to find, we needed to find that direction of the electric field.

02:33 And as i said, since the outer conductor is at a high.

02:38 Higher potential at the inner conductor, the electric field will point inward toward the center axis.

02:45 Now for part b of this problem, we are asked, we are, well, we are tell that the magnitude of the electric field, for part b, the magnitude of the, of the, of the electric field is a constant c over r.

03:04 And it tells us that we need to use the definition for the, electric potential so that we know that the electric potential is defined as the this where the l corresponds to a differential in length so this is the dot product between the electric field and this dl now with this we need to find the value for for the constant c, we need to find the value for the constant c.

03:50 So that's what we are going to do.

03:57 And so we know that definition for the electric field, so we just need simply to substitute those values.

04:05 We can see from the figure that the delt is just a differential in length, and it is it is perpendicular parallel to the vector of the of the electric field.

04:23 This is in red and in black is the electric field and this is the dealt.

04:28 So we can see from that and we note that the definition of the dot product is the integral, okay, the dot product is the magnitude of each of these vectors times the cosine of the angle that it made one with the other.

04:46 So we note that the angle between these two is zero.

04:50 So we will have that cosine of zero.

04:51 Is just one.

04:53 So we'll obtain the magnitude of this, which is this, and the yield that in this case is at the r, because it's a differential in the radius, because we are moving radiantly from the outer shell to the inner shell.

05:10 So we need to integrate from the out.

05:16 We are moving from the, well, the inner shell, you know, we're moving from the inner shell to the r chel.

05:31 And in this case, we substitute the magnitude for the electric field, so it's this.

05:37 And we can take out the constant c because it doesn't depend on art, on the radius.

05:46 So we will have the integral of 1 over r.

05:50 And we know that that integral is the neparian logarithm.

05:53 So we will have c, the logarithm of r...

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