What is the Electric Flux through a Net in a Uniform Electric Field?

[Sunbodi]

Homework Statement

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A butterfly net hangs from a circular loop of diameter 400 mm . You hold the loop horizontally in a region where the electric field is 150 N/C downward, as shown in (Figure 1) .
What is the electric flux through the net?

Homework Equations

k = 9E9 N
E = kq/r²
Flux = Q enclosed / E₀
Flux = EA

The Attempt at a Solution

r = .195 m

150 = k(q)/(.2)²
q = 6.667E-10

Flux = 6.667e-10/8.85E-12
Flux = 75.3

Mastering Physics says this is wrong. Please help?

 

[Sunbodi]

I'm thinking since this is not a closed surface I should instead simply use Flux = E*A so E = the given amount 150 and A = pi*(.2)^2 giving the amount 18.84 for flux.

I think this is correct however I'm down to only 3 tries. Do you think this is instead correct?

 

[gneill]

r = .195 m

150 = k(q)/(.2)²
q = 6.667E-10

This calculation doesn't make any sense to me, since no charge was mentioned. Instead you were given the magnitude and direction of the electric field without any details of the charge distribution that might account for it. You simply need to assume that the field is as given.

I'm thinking since this is not a closed surface I should instead simply use Flux = E*A so E = the given amount 150 and A = pi*(.2)^2 giving the amount 18.84 for flux.

I think this is correct however I'm down to only 3 tries. Do you think this is instead correct?

That would be the correct approach. You want to find the flux passing through the planar surface defined by the loop.

What units will you assign to your flux magnitude of 18.85?

 

[Tanishq Nandan]

Woah,not a closed surface?
Well,then how come you are applying flux=E.A,which comes from Gauss Law,which is only valid for closed surfaces??We are assuming a closed surface(which is in essence the net with a lid on the opening)
Now,since there isn't any charge enclosed in this surface,the total flux passing through this surface ought to be zero.
The rest of the caculation done by you is correct,or rather -18.84 to make the answer perfect.

 

[Sunbodi]

This calculation doesn't make any sense to me, since no charge was mentioned. Instead you were given the magnitude and direction of the electric field without any details of the charge distribution that might account for it. You simply need to assume that the field is as given.

That would be the correct approach. You want to find the flux passing through the planar surface defined by the loop.

What units will you assign to your flux magnitude of 18.85?

N⋅m2/C

Thank you.

 

[Sunbodi]

Woah,not a closed surface?
Well,then how come you are applying flux=E.A,which comes from Gauss Law,which is only valid for closed surfaces??We are assuming a closed surface(which is in essence the net with a lid on the opening)
Now,since there isn't any charge enclosed in this surface,the total flux passing through this surface ought to be zero.
The rest of the caculation done by you is correct,or rather -18.84 to make the answer perfect.

Thank you so much!

 

[gneill]

N⋅m2/C

Yes, those units are adequate. But you're more likely to see electric flux shown with the equivalent units V⋅m .

 

[Sunbodi]

Yes, those units are adequate. But you're more likely to see electric flux shown with the equivalent units V⋅m .

This was before we learned of potential difference. We just learned about voltage (past high school level physics) today. Thanks :)