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Physics Midterm Help Part 2

  • Thread starter Nimmy
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  • #1
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Homework Statement



A point charge 'q' hangs motionless from a mass less string at an angle, due to the forces of gravity and electrical attraction with a nearby infinite conducting plane of charge.

Conducting wire sigma = -3.0 E -8 String is next to the conducting wire with a angle
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a. If the mass of 'q' is 4.00 E -6 kg, and it hangs at angle of 25.0 degrees with the vertical, find the charge 'q'.






Homework Equations



Using the equations, that I learned from mechanics. I broke up the force components in x and y direction.

∑ Fx = Fe - T sin 25º = 0
∑ Fy = T cos 25º - mg = 0


The Attempt at a Solution



∑ Fy = T cos 25º - mg = 0
T = mg/cos 25º
T = (4.00 E-6 kg) (9.80 m/s^2)/cos (25º) = 3.55 x 10^-5 N (I wrote N on my test and he crossed that out)

Electrical Force = T sin 25º = (3.55 x 10^-5N) sin 25º = 1.50 x 10^-5 N

Now I am lost on how to solve for q. Can you pinpoint where Im making errors and also direct me on how to solve for q? Because on my test I got half of the points.

How would you sketch an electric field diagram of 'q' and 'sigma' ? Also, where would the net electric field be exactly zero?
 
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Answers and Replies

  • #2
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Ok for part a....is the process right

so since I already found what Electrical force is. I can just use F = qE ....which means q= F/E ...So I plug in F which I found already and E = Sigma/2 Eplison knott. Since we got the sigma value. But when I solved it I got q = -8.85 x 10^-9 C. Is that right?
 
  • #3
LowlyPion
Homework Helper
3,090
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Homework Statement



A point charge 'q' hangs motionless from a mass less string at an angle, due to the forces of gravity and electrical attraction with a nearby infinite conducting plane of charge.

Conducting wire sigma = -3.0 E -8 String is next to the conducting wire with a angle
| |
| |
| |
| |

a. If the mass of 'q' is 4.00 E -6 kg, and it hangs at angle of 25.0 degrees with the vertical, find the charge 'q'.

Homework Equations



Using the equations, that I learned from mechanics. I broke up the force components in x and y direction.

∑ Fx = Fe - T sin 25º = 0
∑ Fy = T cos 25º - mg = 0


The Attempt at a Solution



∑ Fy = T cos 25º - mg = 0
T = mg/cos 25º
T = (4.00 E-6 kg) (9.80 m/s^2)/cos (25º) = 3.55 x 10^-5 N (I wrote N on my test and he crossed that out)

Electrical Force = T sin 25º = (3.55 x 10^-5N) sin 25º = 1.50 x 10^-5 N

Now I am lost on how to solve for q. Can you pinpoint where Im making errors and also direct me on how to solve for q? Because on my test I got half of the points.

How would you sketch an electric field diagram of 'q' and 'sigma' ? Also, where would the net electric field be exactly zero?
Draw a diagram and I would think you would discover that the ratio of the Forces - Gravity down to Electrostatic to the E-Field will be given by

Felectro / Fgravity = Tan25
 
  • #4
LowlyPion
Homework Helper
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Ok for part a....is the process right

so since I already found what Electrical force is. I can just use F = qE ....which means q= F/E ...So I plug in F which I found already and E = Sigma/2 Eplison knott. Since we got the sigma value. But when I solved it I got q = -8.85 x 10^-9 C. Is that right?
F = qE

should deliver the right answer if you have the proper F.
 
  • #5
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F = qE

should deliver the right answer if you have the proper F.
So the E is E = sigma/2 Epslion knott?
 
  • #6
LowlyPion
Homework Helper
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So the E is E = sigma/2 Epslion knott?
Conducting plane or conducting wire?
 
  • #7
LowlyPion
Homework Helper
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I don't understand this statement:
Conducting wire sigma = -3.0 E -8 String is next to the conducting wire with a angle
If it is a conducting plane your formula would be correct.
 
  • #8
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I don't understand this statement:


If it is a conducting plane your formula would be correct.
Yup a conducting plane of charge. I was trying to explain the diagram. Thanks though! :)

I have one question is the net electric field zero at the center of the conducting plane?
 
  • #9
LowlyPion
Homework Helper
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Yup a conducting plane of charge.
OK. That should be good to go then.

And the conducting wire is from what? Did you mean to write plane? Because if you have a wire you need to know r from the wire, which I don't see in the problem as given.
 
  • #10
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OK. That should be good to go then.

And the conducting wire is from what? Did you mean to write plane? Because if you have a wire you need to know r from the wire, which I don't see in the problem as given.
Yes there is no conducting wire....it's a conducting plane of charge. Plus there was no radius given. But can you answer where the net electric field is zero in this problem? B/C I though it was at the center. But I got ding -10 points on the exam.
 
  • #11
LowlyPion
Homework Helper
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Yes there is no conducting wire....it's a conducting plane of charge. Plus there was no radius given. But can you answer where the net electric field is zero in this problem? B/C I though it was at the center. But I got ding -10 points on the exam.
The E field is 0 inside the conductor.

As to the E-field, I think I misspoke. I think it should be σ / εo
see:
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/gausur.html#c3
 
  • #12
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  • #13
LowlyPion
Homework Helper
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so it just should be 1/2 of the value that I got before right? Oh is the E field zero inside the conductor because the E field go in and out of the conductor cancel out at the center?
Wouldn't it be twice?

Didn't you say σ /2 εo ?

That would be for a capacitor I think.

In a conductor the charge is at the surface. Inside the conductor any Gaussian surface you draw that does not contain the surface must result in a 0 field, because there would be no charge within the closed surface.
 
  • #14
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Wouldn't it be twice?

Didn't you say σ /2 εo ?

That would be for a capacitor I think.


In a conductor the charge is at the surface. Inside the conductor any Gaussian surface you draw that does not contain the surface must result in a 0 field, because there would be no charge within the closed surface.
Well, because I got 1.50 E -5 N for the Electrical Force. sigma = -3.0 E -8 C/m^2)
Now if it was 2 εo

q= F/E = (1.50 E -5 N)/(-3.0 E -8 C/m^2 /2(8.85 E -11 N/m) = -8.85 E -9 C

Now if The E Field = σ / εo

So ...q =F/E = (1.50 E -5 N)/ (-3.0 E -8 C/m^2 /(8.85 E -11 N/m) = -4.43 E -9 C

Thats why I said half the value.
 

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