Electric flux through an inclined cylinder

[Zayan]

Homework Statement

Finding the electric flux entering a cylinder(height H radius R) in uniform Electric Field. Axis of cylinder inclined at an angle theta from vertical.

Relevant Equations

Flux = E.A

Flux equals dot product of Area Vector and the Field vector. So the areas are 3. The projection of cylinder (shadow of cylinder). It's flux is E2HRcostheta. Then remaining two circles have flux Eπr²sintheta each. But I checked my solution book and they have considered only half areas of the top and bottom circle. Why is that?

 

[Orodruin]

Please give the problem statement exactly as provided. The most straightforward interpretation of what you wrote here is trivially zero.

 

[Zayan]

Please give the problem statement exactly as provided. The most straightforward interpretation of what you wrote here is trivially zero.

It would be zero if it was the net flux. I stated that it is asking about the entering flux.

 

[TSny]

Flux equals dot product of Area Vector and the Field vector. So the areas are 3. The projection of cylinder (shadow of cylinder). It's flux is E2HRcostheta. Then remaining two circles have flux Eπr²sintheta each. But I checked my solution book and they have considered only half areas of the top and bottom circle. Why is that?

Carefully inspect the field lines. For example, look at the 4th field line from the bottom. It pierces the bottom circle and also pierces the light- colored flat region of area 2RH (axial plane) in your second figure.

 

[Zayan]

Carefully inspect the field lines. For example, look at the 4th field line from the bottom. It pierces the bottom circle and also pierces the light- colored flat region of area 2RH (axial plane) in your second figure.

Do you mean like, some of the area is already included in the flat region that's why we don't take it?

 

[TSny]

Do you mean like, some of the area is already included in the flat region that's why we don't take it?

Yes. You need to avoid double counting some of the flux.

Also, note that all of the field lines that pierce the top circle are coming out of the cylinder, not into the cylinder.

 

[Zayan]

Yes. You need to avoid double counting some of the flux.

Also, note that all of the field lines that pierce the top circle are coming out of the cylinder, not into the cylinder.

Yeah I got it. Since I took the flux of the projection of the axis of cylinder, it already has common area with half the top circle and half the bottom circle. That's why we only need other halves.

 

[Orodruin]

It would be zero if it was the net flux. I stated that it is asking about the entering flux.

That’s ambiguous. Many authors will count exiting flux as negative entering flux.

 

[Orodruin]

As for the problem as you wish it - just compute the projection of the cylinder on a plane perpendicular to the field and multiply by the field strength.

 

[kuruman]

In other words, find the area of the shadow that the cylinder casts on a plane perpendicular to the field and multiply it by the value of the field.

 

[Zayan]

In other words, find the area of the shadow that the cylinder casts on a plane perpendicular to the field and multiply it by the value of the field.

I already did that. This is an inclined cylinder so there are 3 areas, projection, top circle and bottom circle. Since the projection covers half the top and half the bottom circle we only include one complete circular area of flux hence the discrepancy.

 

[kuruman]

I already did that. This is an inclined cylinder so there are 3 areas, projection, top circle and bottom circle.

What circle?

See the figure on the right. You have two halves of an ellipse of semi-major axis $a=R$ and semi-minor axis $b=R\sin\!\theta$ and a $(2R\times H\cos\!\theta)$ rectangle.

 

[Orodruin]

What circle?

While the end caps’ projection on a plane are ellipses, they are circles in three dimensions, which is most likely what the OP refers to.

 

[kuruman]

It looks like the solution provided $$

While the end caps’ projection on a plane are ellipses, they are circles in three dimensions, which is most likely what the OP refers to.

I suppose it is a language problem. OP does mention

remaining two circles have flux Eπr²sintheta each

which, taken at face value, implies that each "circle" has radius $r\sqrt{\sin\!\theta.}$ I don't think that OP has correctly thought through how the flat circular faces project onto the plane.

 

[Orodruin]

which, taken at face value, implies that each "circle" has radius rsinθ. I don't think that OP has correctly thought through how the flat circular faces project onto the plane.

… or that the circle has radius r and the field has a normal component $E\sin\theta$, which is the actual case.

 

[kuruman]

… or that the circle has radius r and the field has a normal component $E\sin\theta$, which is the actual case.

Yes.

 

[TSny]

For fun, another way to the answer:

Region $A$ is the “axial plane” shown in the second figure of the OP. Regions $B, C$, and $D$ are semicircles. Region $F$ is the left half of the lateral side of the cylinder.

The total flux into the cylinder is $$|\Phi_{tot}^{in}| = |\Phi_F|+|\Phi_B|+|\Phi_C|$$
Consider the closed surface consisting of $A + D + F + B$. This surface encloses no charge, so the flux into the surface equals the flux out: $$|\Phi_F| + |\Phi_B| = |\Phi_A| + |\Phi_D|.$$ Using this in the first equation: $$|\Phi_{tot}^{in}| = |\Phi_A|+|\Phi_D|+|\Phi_C| = 2ERH \cos \theta + E\frac{\pi R^2}{2} \sin \theta + E\frac{\pi R^2}{2} \sin \theta.$$