[Griffiths Electrodynamics] Finding the electric flux through this quarter cylinder

[physics_student01]

TL;DR Summary: Confusion in φ cap direction for Left & Back face

The image of the question is shared at this Imgur link - The only issue I'm facing in this particular problem is in calculating the flux through the "Left" & "Back" faces of this quarter cylinder.

More specifically the issue is in the direction of φ cap, because I'm unable to understand or figure out how or why we considered it to be negative for Left face & positive for the Back face. This is something I'm unable to visualise. Further adding fuel to the fire is the knowledge that unlike Cartesian system, these coordinates don't have a "fixed" spatial direction as such. So how can & how are we assigning positive-negative directions here?

Additionally, it still didn't make sense for the negative angle direction except when I consider this- φ is the angle made wrt the x axis. For the left face, the direction is negative because the angle makes a clockwise movement. In case of the back face however, the direction is positive because the angle traced while reaching that particular face is anti-clockwise. Does it make sense to you too/is my interpretation correct?

Please guide.

 

[kuruman]

Which problem are you asking about, 1.42 or 1.43? By convention, $\mathbf{\hat {\phi}}$ points in the direction of increasing $\phi$ which is anti-clockwise looking down the $z-$axis.

 

[TSny]

More specifically the issue is in the direction of φ cap, because I'm unable to understand or figure out how or why we considered it to be negative for Left face & positive for the Back face.

When finding the flux $\oint \mathbf v \cdot \mathbf{da}$ of a vector field $\mathbf v$ through a closed surface, the area vectors $\mathbf{da}$ point outward from the enclosed region. So, you need to consider the relation between $\mathbf{da}$ and $\hat{\boldsymbol{\phi}}$ for the left and back sides of the quarter-cylinder.

 

[hutchphd]

Care must be taken calculating flux from a closed surface. The efflux (from inside to outside) is defined as positive (this also coincides with the definition of the unit surface normal vector). So you need to be cognizant and adjust the calculation appropriately.

 

[physics_student01]

Which problem are you asking about, 1.42 or 1.43? By convention, $\mathbf{\hat {\phi}}$ points in the direction of increasing $\phi$ which is anti-clockwise looking down the $z-$axis.

Thank you for responding! My issue is with 1.42, one of its parts where Flux needs to be calculated.

I believe as per your insight, my interpretation mentioned in #1 seems correct. Do you also think so?

 

[physics_student01]

When finding the flux $\oint \mathbf v \cdot \mathbf{da}$ of a vector field $\mathbf v$ through a closed surface, the area vectors $\mathbf{da}$ point outward from the enclosed region. So, you need to consider the relation between $\mathbf{da}$ and $\hat{\boldsymbol{\phi}}$ for the left and back sides of the quarter-cylinder.

Yes indeed. But for the Left & Back side, somehow it is not clear why one is positive & the other negative. I was suspecting it has to do with clockwise/anticlockwise sweep of the angle as mentioned in #1. Is that a correct way of thinking about it?

 

[hutchphd]

Is that a correct way of thinking about it?

The issue is the sign of $\vec v \cdot d\vec a$ as explained above

When finding the flux ∮v⋅da of a vector field v through a closed surface, the area vectors da point outward from the enclosed region. So, you need to consider the relation between da and ϕ^ for the left and back sides of the quarter-cylinder.

I don't see how it can be said more clearly

 

[Brahim0573]

To find the flux ∮v⋅da of a vector field v through a closed surface, such as a quarter-cylinder, it's crucial to consider the orientation of the area vectors da with respect to the direction of the vector field v.
Let's consider the quarter-cylinder with its curved surface and two flat surfaces (front and bottom) forming the closed surface. When we talk about the orientation of the area vectors da, we mean their direction relative to the outward normal of the surface.
For the curved surface of the quarter-cylinder, the outward normal points radially outward from the cylinder. Therefore, the orientation of the area vector da on this surface would also be radially outward, perpendicular to the surface.
For the flat surfaces (front and bottom), the outward normal is straightforward, perpendicular to the surface and pointing away from the enclosed region. Therefore, the orientation of the area vectors da on these surfaces would be aligned with the outward normal.
In terms of the relation between da and the unit vector ϕ^ϕ^ (which typically denotes the azimuthal direction in cylindrical coordinates), we can say the following:

• For the curved surface of the quarter-cylinder, since the orientation of da is radial, and in cylindrical coordinates, ϕ^ϕ^ represents the azimuthal direction which is tangent to the circular cross-sections of the cylinder, there's no direct relation between da and ϕ^ϕ^ on this surface.
• For the flat surfaces (front and bottom), since da is oriented perpendicular to the surface and ϕ^ϕ^ points tangentially to the circular cross-sections, there's no direct relation between da and ϕ^ϕ^ on these surfaces either.

However, while integrating over the entire surface to find the flux, you will need to consider the dot product v⋅dav⋅da at each point on the surface, taking into account the orientation of both the vector field v and the area vector da at that point. This involves aligning the direction of v and da appropriately to calculate the dot product correctly, ensuring that the flux is calculated accurately.

 

[hutchphd]

The finite length quarter cylindar has 5 surfaces: four of them are flat. Have you forgotten the surfaces in the x-z plane and in the y-z plane???? What about them?

 

[berkeman]

Thread locked temporarily -- @physics_student01 check your messages and please reply to my latest message with the question about the other account. Thanks.