Electric Flux through a circle

In summary, the conversation discusses the concept of a Gaussian surface and its relation to a circular disc bounded by a circle. The process of solving a problem using a surface integral is also mentioned. The conversation then moves on to solving part b of the problem, using a spherical coordinate system and calculating the integral of the dot product between the electric field and the surface element. The expert provides guidance and clarifications throughout the conversation.
  • #1
424
14
Homework Statement
A point charge q is located at the origin. Consider the electric field
flux through a circle a distance from q, subtending an angle 2θ, as
shown in Fig. 1.36. Since there are no charges except at the origin,
any surface that is bounded by the circle and that stays to the right
of the origin must contain the same flux. (Why?) Calculate this
flux by taking the surface to be:

(a) the flat disk bounded by the circle;
(b) the spherical cap (with the sphere centered at the origin)
bounded by the circle.
Relevant Equations
Gauss Law
fig.PNG


Hi! My main problem is that I don't understand what the problem is telling me. What does it mean that the surface is a flast disc bounded by the circle? Is the Gauss surface the disc? Does that mean that inside the circle in the figure, there is a disc?

Can you give me some guidance on how to begin solving the problem?
 
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
circle.png

The black (and curved) line is the circle. The blue area is the disc bounded by the circle.
A circular disc is a surface but is not a gaussian surface. A gaussian surface is a closed surface that encloses a 3D volume pretty much like the surface of a sphere encloses the volume of the sphere. if you can imagine that the blue disc blows and stretches (while the boundary of the disc (the black circle) remains unstretched) towards yourself then this streched and blown disc would make a gaussian surface( I know its kind of hard to explain with words without making 3D schemes).
To start solving a) you must setup a surface integral $$\int\int_A\vec{E}\cdot\vec{dS}$$ where E is the expression for the electric field in the various points of the circular disk and A is the set of points of space that makeup the circular disk. Then you got to calculate the integral.
 
  • #3
Delta2 said:
View attachment 249473
The black (and curved) line is the circle. The blue area is the disc bounded by the circle.
A circular disc is a surface but is not a gaussian surface. A gaussian surface is a closed surface that encloses a 3D volume pretty much like the surface of a sphere encloses the volume of the sphere. if you can imagine that the blue disc blows and stretches (while the boundary of the disc (the black circle) remains unstretched) towards yourself then this streched and blown disc would make a gaussian surface( I know its kind of hard to explain with words without making 3D schemes).
To start solving a) you must setup a surface integral $$\int\int_A\vec{E}\cdot\vec{dS}$$ where E is the expression for the electric field in the various points of the circular disk and A is the set of points of space that makeup the circular disk. Then you got to calculate the integral.
I understand, thanks. I think I made it, is this correct?
20190911_081823.jpg
20190911_081854.jpg
 
  • #4
Work at image1 looks good.

But I don't understand why in your subsequent work in image2 you move the ##\cos^3\theta## term out of the integral...
 
  • #5
Delta2 said:
Work at image1 looks good.

But I don't understand why in your subsequent work in image2 you move the ##\cos^3\theta## term out of the integral...
I didn't notice, sorry. I corrected and end up with this, I think it is correct.
20190911_092607.jpg
 
  • #7
Yes I think it is correct.

For b) use a spherical coordinate system such that the inclination (or polar) angle will be the ##\theta## of figure 1.36 (we basically work in the same coordinate system as in image 1, regarding ##\phi## and ##\theta##, and ##R## now is the radius of the sphere centered at origin). The surface element of the spherical cap is ##dS=R^2\sin\theta d\theta d\phi## in that system. The normal to the surface element of the spherical cap will be always in the same direction as the direction of the electric field in that surface element. So the dot product simplifies to $$\vec{E}\cdot d\vec{S}=|\vec{E}||d\vec{S}|=|\frac{q}{4\pi\epsilon_0R^2}||R^2\sin\theta| d\theta d\phi$$. So all you have to calculate is the integral of the above.
 
Last edited:
  • #8
Delta2 said:
Yes I think it is correct.

For b) use a spherical coordinate system such that the inclination (or polar) angle will be the ##\theta## of figure 1.36. In the case of the spherical cap, the normal to the surface element of the spherical cap will be always in the same direction as the direction of the electric field in that surface element. So the dot product simplifies to ##\vec{E}\cdot d\vec{S}=|\vec{E}||d\vec{S}|=|\frac{q}{4\pi\epsilon_0R^2}||R^2\sin\theta| d\theta d\phi##. So all you have to calculate is the integral of the above.
I see, thank you! You've helped me a lot.
 
  • Like
Likes Delta2

What is electric flux through a circle?

Electric flux through a circle is the measure of the total electric field passing through a circular surface. It is calculated by taking the dot product of the electric field and the area vector of the circle.

How is electric flux through a circle calculated?

Electric flux through a circle is calculated by taking the dot product of the electric field and the area vector of the circle. The formula is Φ=∫E⃗ ⋅ dA⃗, where Φ is electric flux, E⃗ is the electric field, and dA⃗ is the area vector of the circle.

What is the unit of electric flux through a circle?

The unit of electric flux through a circle is volts per meter (V/m) or, equivalently, newtons per coulomb (N/C).

What factors affect the electric flux through a circle?

The electric flux through a circle is affected by the strength of the electric field, the size and orientation of the circle, and the angle between the electric field and the normal vector of the circle.

How is the direction of electric flux through a circle determined?

The direction of electric flux through a circle is determined by the direction of the electric field and the orientation of the circle's area vector. The flux will be positive if the electric field and the area vector are in the same direction, and negative if they are in opposite directions.

Suggested for: Electric Flux through a circle

Replies
6
Views
214
Replies
28
Views
2K
Replies
17
Views
6K
2
Replies
38
Views
2K
Back
Top