What is the strength of the electric field 0.1 mm above the

[itzernie]

Homework Statement

[/B]
A thin, horizontal, 11-cm-diameter copper plate is charged to 4.3 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Homework Equations

[/B]
E= n/(e₀)
n = Q/A
E = Q/(2Ae₀)
Q = 4.3 E -9
e₀ = 8.85 E-12
A = π(.055)^2 = 9.5 E -3

The Attempt at a Solution

I tried plugging in the info provided to solve for E. I placed a 2 in the denominator since the problem is only dealing with one plate instead of 2. by doing so I get 25572.41 or 2.6 E 4 since the problems calls for 2 sig figs. This number does not include the 1mm distance from the surface which is what i am confused on. I tried multiplying by 1 E-3 but that did not work

 

[Student100]

Homework Statement

[/B]
A thin, horizontal, 11-cm-diameter copper plate is charged to 4.3 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?

Homework Equations

[/B]
E= n/(e₀)
n = Q/A
E = Q/(2Ae₀)
Q = 4.3 E -9
e₀ = 8.85 E-12
A = π(.055)^2 = 9.5 E -3

The Attempt at a Solution

I tried plugging in the info provided to solve for E. I placed a 2 in the denominator since the problem is only dealing with one plate instead of 2. by doing so I get 25572.41 or 2.6 E 4 since the problems calls for 2 sig figs. This number does not include the 1mm distance from the surface which is what i am confused on. I tried multiplying by 1 E-3 but that did not work

Plugged the info into what exactly? $$\frac{\sigma}{2\epsilon_0}$$??

And you put a two so $$\frac{\sigma}{4\epsilon_0}$$??

Can you clear up your approach?

 

[itzernie]

Sure! Sorry about that. From lecture notes I copied down the formula E = σ/ε₀ to find the field if two plates were involved. Our professor was then saying if only one plate was involved we should use E = σ/2ε₀ , which is what I used.

Then σ = Q/A
Q being charge
A being area

so now our equation, subbing this into E, looks like E = Q/2Aε₀
Q is given, A is solved for using 2πr² , and ε₀ is a constant.

 

[Student100]

Sure! Sorry about that. From lecture notes I copied down the formula E = σ/ε₀ to find the field if two plates were involved. Our professor was then saying if only one plate was involved we should use E = σ/2ε₀ , which is what I used.

Then σ = Q/A
Q being charge
A being area

so now our equation, subbing this into E, looks like E = Q/2Aε₀
Q is given, A is solved for using 2πr² , and ε₀ is a constant.

That would be correct then, just wanted to make sure you weren't multiplying by 4. The other question you have comes from not deriving the result yourself, the distance and symmetry of the test point above the circular plate is actually what allows you to use the above. Have you derived the result for a force on a charge using disks and symmetry before? It's along the same lines.

Have you ever seen: $$E_z = \frac{\sigma}{2\epsilon_0}(1-\frac{z}{\sqrt{z^2+r^2}})$$ Where we're looking at the electric field at a point on the z axis?

 

[itzernie]

I do not believe we have done that yet, but I believe it is coming up next week (unfortunately after the homework is due :( ) As for force, the only info I really have is that F = qE, not sure if that helps in this case

 

[itzernie]

Not sure if I missed it the first time, but I recently noticed the equation you just placed into reply #4. I have actually never seen that yet, but by the looks of it it has pieces that makes sense. r would be .055m and z would be the distance off of the surface ? Also, the z/√(z ²+ r²) ... that looks like it could be the sin/ cos of an angle ?

Thanks for all of this help by the way!

 

[Student100]

I do not believe we have done that yet, but I believe it is coming up next week (unfortunately after the homework is due :( ) As for force, the only info I really have is that F = qE, not sure if that helps in this case

Okay, well you'll actually look at the limiting case as r -> infinity, which it turns out is useful for when you're very close to a uniformly charged surface and let's you use the above. Since you're close to the disk in this case compared to it's size, and above it's center, you can use the above.

Also I just caught it, but the area is just $\pi r^2$ for the above.

 

[Student100]

Not sure if I missed it the first time, but I recently noticed the equation you just placed into reply #4. I have actually never seen that yet, but by the looks of it it has pieces that makes sense. r would be .055m and z would be the distance off of the surface ? Also, the z/√(z ²+ r²) ... that looks like it could be the sin/ cos of an angle ?

Thanks for all of this help by the way!

Yes, you're just using the approximate result that's already been simplified. You'll see when you do the derivation in class hopefully, if you don't do it, and it's a calculus based course, something is wrong.

 

[itzernie]

I just recalculated everything and I got the correct answer. Thanks ! There was also another question for 1mm below which I realized would be the same as 1mm above due to symmetry I believe. Thank you for all of your help!

 

[Student100]

I just recalculated everything and I got the correct answer. Thanks ! There was also another question for 1mm below which I realized would be the same as 1mm above due to symmetry I believe. Thank you for all of your help!

Yes, just keep in mind direction as its a vector.