Field dependence on the potential in a wire

[Krushnaraj Pandya]

Homework Statement

Does a point in a wire having higher potential necessarily mean that it has a higher electric field?
For example consider the cases of question 27 and 31 in the picture attached below.

Homework Equations

E=V/L

The Attempt at a Solution

The potential at a and b is the same in 27, so can I say E is the same. What does L in E=V/L mean here in this context? In 31, I think E at a is higher since potential is higher, still not sure about what L means

 

[gneill]

Can't read the text in the picture. Please type out the text content and provide close-ups of the figures

 

[Krushnaraj Pandya]

Can't read the text in the picture. Please type out the text content and provide close-ups of the figures

really sorry about that.
consider a cylindrical resistor whose resistivity increases linearly with the radial distance. it is connected to a battery as shown. (Now refer figure 1), If the electric fields at a and b are Ea and Eb then how are they related?
Now consider a thick cylinder of uniform resistivity. A battery is connected as shown. Same question, (see figure 2)

 

[haruspex]

Does a point in a wire having higher potential necessarily mean that it has a higher electric field?

No. The field goes with the potential gradient, not the potential value.

 

[Krushnaraj Pandya]

No. The field goes with the potential gradient, not the potential value.

So Edx=dV...but how do we compare V/l in the context of 31?

 

[Delta2]

For 31. the current densities at A and B are (from ohm's law) $J_A=\frac{E_A}{\rho}$ and $J_B=\frac{E_B}{\rho}$ since the resistivity $\rho$ is the same. The total current through the collateral surface at A $S_A$ must be equal to the current through the collateral surface at B $S_B$. So we have

$J_AS_A=J_BS_B$

From this equation you can figure out which of $J_A, J_B$ is bigger (depending which of $S_A, S_B$ is bigger ) and hence which of the electric field is bigger.

 

[Krushnaraj Pandya]

For 31. the current densities at A and B are (from ohm's law) $J_A=\frac{E_A}{\rho}$ and $J_B=\frac{E_B}{\rho}$ since the resistivity $\rho$ is the same. The total current through the collateral surface at A $S_A$ must be equal to the current through the collateral surface at B $S_B$. So we have

$J_AS_A=J_BS_B$

From this equation you can figure out which of $J_A, J_B$ is bigger (depending which of $S_A, S_B$ is bigger ) and hence which of the electric field is bigger.

From this Ja is greater and therefore Ea is greater. But by the same logic, in 27- area that charge passes through is the same for A and B therefore Ja=Jb and since resistivity increases linearly with radius, Eb should be greater than Ea which is incorrect.
Btw, I read your insight on the fundamental theorem of calculus and I really enjoyed whatever little I could grasp from it. :D

 

[Delta2]

Hard to explain without making a schematic, but that equation with current densities and surfaces does not hold in the case of 27. Remember that equation comes from the condition that the current through different surfaces that are connected in a series-way must be equal. It is also because the vector AB is perpendicular to the direction of the current flow , while in 31 the vector AB is parallel to the direction of the current flow.

 

[Krushnaraj Pandya]

Hard to explain without making a schematic, but that equation with current densities and surfaces does not hold in the case of 27. Remember that equation comes from the condition that the current through different surfaces that are connected in a series-way must be equal. It is also because the vector AB is perpendicular to the direction of the current flow , while in 31 the vector AB is parallel to the direction of the current flow.

Is there any intuitive way to be able to find that Ea=Eb in 27? If it's beyond what I can currently understand then we can do away with mathematical rigor- but I feel like if that question is in my book there must be simple way to understand this

 

[Delta2]

Is there any intuitive way to be able to find that Ea=Eb in 27? If it's beyond what I can currently understand then we can do away with mathematical rigor- but I feel like if that question is in my book there must be simple way to understand this

Yes the simple intuitive way is to view it as a capacitor with electric field E=V/L between the two plates (left and right cap of the cylinder) where L is the length of the cylinder. The difference of course is that there is current between the two plates (while in a capacitor there is no conventional current that is flow of charges between the two plates) but because the direction of current is axial , that is perpendicular to the direction of the variable resistivity which is radial, the variable resistivity doesn't affect the electric field.

 

[Krushnaraj Pandya]

Yes the simple intuitive way is to view it as a capacitor with electric field E=V/L between the two plates (left and right cap of the cylinder) where L is the length of the cylinder. The difference of course is that there is current between the two plates (while in a capacitor there is no conventional current that is flow of charges between the two plates) but because the direction of current is axial , that is perpendicular to the direction of the variable resistivity which is radial, the variable resistivity doesn't affect the electric field.

Got it! Thank you