Electric Flux through half a cylinder

[jlmccart03]

Homework Statement

In (Figure 1) take the half-cylinder's radius and length to be 3.4 cm and 15 cm, respectively.

If the electric field has magnitude 5.3 kN/C, find the flux through the open half-cylinder. The surface here is the right half of the surface of a full cylinder. The surface does not include the rectangle which is the opening to the half-cylinder. Hint: You don't need to do an integral! Why not?

Homework Equations

φ = Qenc / ε₀ = ∫E⋅dA = EA

The Attempt at a Solution

So I simply took my flux equation to be EA since its a half cylinder with constant E and symmetry. Through this I got 5.3 kN/C * (π(3.4cm)²+π(3.4cm)(15cm)) = 1041.65 kN/C m². This is apparently wrong? What is proper way of handling this problem as I though I was doing it correctly.

 

[gneill]

The problem is designed to get you to recognize that the flux that passes through the curved surface of the half-cylinder is the same as the flux that passes through the open rectangular side of the half-cylinder. It's much easier to work with a rectangular area that's perpendicular to the field!

If you choose to calculate using the curved surface then you'll need to do the integration and take into account the direction of the dA element vectors in relation to the field lines and use the dot product (E⋅dA = |E| |dA| cos(φ)) for the contribution for each differential area element.

 

[jlmccart03]

The problem is designed to get you to recognize that the flux that passes through the curved surface of the half-cylinder is the same as the flux that passes through the open rectangular side of the half-cylinder. It's much easier to work with a rectangular area that's perpendicular to the field!

If you choose to calculate using the curved surface then you'll need to do the integration and take into account the direction of the dA element vectors in relation to the field lines and use the dot product (E⋅dA = |E| |dA| cos(φ)) for the contribution for each differential area element.

Wait since the open rectangular side is perpendicular to the E field doesn't that mean the flux is 0?
EDIT That is wrong. I don't know what I was thinking. So i just have to take E * Rectangle Area = Flux.

 

[gneill]

Wait since the open rectangular side is perpendicular to the E field doesn't that mean the flux is 0?
EDIT That is wrong. I don't know what I was thinking. So i just have to take E * Rectangle Area = Flux.

Bingo!

 

[jlmccart03]

Bingo!

Ok so I get 270.3 kN/C * cm^2. I took 5.3 kN/C * (3.4 cm)(15 cm) to get this, but it says that is wrong. I then tried to convert everything into normal units of N/C m^2 and got 27.03 N/C m^2 and that is wrong too. What is the issue? Rectangle area is length * width so I don't get what is wrong.

 

[gneill]

Can you show your calculations with conversions in detail? I'd like to see the intermediate values.

 

[jlmccart03]

Can you show your calculations with conversions in detail? I'd like to see the intermediate values.

Sure. (3.4 cm)(15 cm) = 51 cm. (5.3 kN/C)(51 cm) = 270.3 kN/C cm^2. Also I just realized that I am using 3.4 cm which is the radius of the cylinder am I supposed to use that?
EDIT I figured out that the side is length 6.8 cm so my new answer should be 540.6 kN/C cm^2.

 

[gneill]

Sure. (3.4 cm)(15 cm) = 51 cm. (5.3 kN/C)(51 cm) = 270.3 kN/C cm^2. Also I just realized that I am using 3.4 cm which is the radius of the cylinder am I supposed to use that?

Only if you want to find only half the area

 

[jlmccart03]

Only if you want to find only half the area

Ok, so with 6.8 cm as the new full diameter then I should get 540.6 kN/C cm^2. Now the next question is. Are those appropriate units? Should I have it as 54.06 N/C m^2?

 

[gneill]

Ok, so with 6.8 cm as the new full diameter then I should get 540.6 kN/C cm^2. Now the next question is. Are those appropriate units? Should I have it as 54.06 N/C m^2?

It's always best to convert to the basic units. So N, C, m,...

 

[jlmccart03]

It's always best to convert to the basic units. So N, C, m,...

Noted! Thanks!