Electric Flux
Electric Flux
We have used electric field lines to visualize electric fields and indicate their strength.
We are now going to count* the number of electric field lines passing through a surface, and use this count to determine the electric field.
E
*There are 3 kinds of people in this world: those who can count, and those who can’t.
The electric flux passing through a surface is the number of electric field lines that pass through it.
Because electric field lines are drawn arbitrarily, we quantify electric flux like this: ΦE=EA,
…except that…
If the surface is tilted, fewer lines cut the surface.
E
A
Later we’ll learn about magnetic flux, which is why I will use the subscript E on electric flux.
E
θ
The green lines miss!
E
θ
θ
A
The “amount of surface” perpendicular to the electric field is A cos θ.
AEffective = A cos θ so ΦE = EAEffective = EA cos θ.
We define A to be a vector having a magnitude equal to the area of the surface, in a direction normal to the surface.
Therefore, the amount of surface area effectively “cut through” by the electric field is A cos θ.
Remember the dot product from Physics 1135?
If the electric field is not uniform, or the surface is not flat…
divide the surface into infinitesimal surface elements and add the flux through each…
dA
E
ΔA
Remember, the direction of dA is normal to the surface.
a surface integral, therefore a double integral
If the surface is closed (completely encloses a volume)…
E
…we count* lines going out as positive and lines going in as negative…
dA
a surface integral, therefore a double integral
For a closed surface, dA is normal to the surface and always points away from the inside.
∫
∫
What the *!@* is this thing?
Nothing to panic about!
The circle just reminds you to integrate over a closed surface.
Question: you gave me five different equations for electric flux. Which one do I need to use?
Answer: use the simplest (easiest!) one that works.
Flat surface, E || A, E constant over surface. Easy!
Flat surface, E not || A, E constant over surface.
Flat surface, E not || A, E constant over surface.
Surface not flat, E not uniform. Avoid, if possible.
Closed surface.
If the surface is closed, you may be able to “break it up” into simple segments and still use ΦE=E·A for each segment.
This is the definition of electric flux, so it is on your equation sheet.
The circle on the integral just reminds you to integrate over a closed surface.
A note on terminology…
For our purposes, a vector is constant if its magnitude and direction do not change with position or time.
The electric field is a vector field, so a constant electric field is one that does not change with position or time.
Because the electric field can extend throughout space, we use the term “uniform electric field” to describe an electric field that is constant everywhere in space and time.
A “uniform electric field” is like a “frictionless surface.” Useful in physics problems, difficult (impossible?) to achieve in reality.
In Physics 2135, you can use the terms “constant electric field” and “uniform electric field” interchangeably.
Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.
E
To be worked at the blackboard in lecture…
Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.
E
I see three parts to the cylinder:
The left end cap.
E
dA
Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.
E
I see three parts to the cylinder:
The tube.
E
Electric Flux Example: Calculate the electric flux through a cylinder with its axis parallel to the electric field direction.
E
I see three parts to the cylinder:
The right end cap.
E
dA
Let’s separately calculate the contribution of each part to the flux, then add to get the total flux.
The right end cap.
E
dA
The tube.
E
E
dA
The left end cap.
E
dA
The left end cap.
Every dA on the left end cap is antiparallel to E. The angle between the two vectors is 180°
E is uniform, so
The tube.
E
Let’s look down the axis of the tube.
E is pointing at you.
◉
Every dA is radial (perpendicular to the tube surface).
dA
The angle between E and dA is 90°.
dA
E
E
The angle between E and dA is 90°.
dA
E
The tube contributes nothing to the flux!
Every dA on the right end cap is parallel to E. The angle between the two vectors is 0°
E is uniform, so
The right end cap.
E
dA
The net (total) flux
The flux is zero! Every electric field line that goes in also goes out.
Assuming a right circular cylinder.*
Thanks