# Physics Midterm Help Part 2

## 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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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?

LowlyPion
Homework Helper

## 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

LowlyPion
Homework Helper
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.

F = qE

should deliver the right answer if you have the proper F.

So the E is E = sigma/2 Epslion knott?

LowlyPion
Homework Helper
So the E is E = sigma/2 Epslion knott?
Conducting plane or conducting wire?

LowlyPion
Homework Helper
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.

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?

LowlyPion
Homework Helper
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.

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.

LowlyPion
Homework Helper
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

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

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?

LowlyPion
Homework Helper
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.

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.