Solving Spherical Conductor Homework: E, q, r2, C, ε₀

In summary: I don't know the answer to part#3.In summary, a spherical conductor has an Electric field near its surface of 180 volts per inch. The sphere has a total charge of 7088.61 coulombs. The sphere will not produce an arc because the air around it will not break down.
  • #1
jayz618
27
0

Homework Statement



A Spherical conductor has an Electric field near its surface of 180 volts per inch.
How much charge (in Coulombs) is on the sphere ?
How many excess electrons are on the sphere?
Will the air around the sphere break down and produce an arc ?

Homework Equations



Well, I assume Gauss Law is going to be used

E= Ke ( q / r2)

isolated sphere:
C = 4πε₀r in Farads
ε₀ is 8.8542e-12 F/m
r is radius in m

The Attempt at a Solution



Well, I haven't really been able to start getting anywhere with it, but a few things I know.

That once I am able to determine the total charge of the Sphere, I will be able to calculate the amount of excess electrons with simple math.

It seems to me that we would need to know the size of the sphere in this situation.
 
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  • #2
well, you can answer #3 already, because the E-field is given.
You're right that the _numerical_ answer to #1 will depend on the size of the ball.
They might want you to derive a _formula_ for Q , as a function of R ;
Physics 2 questions do this a LOT more often than Physics 1 questions .
on the other hand, that would make #2 quite peculiar.
 
  • #3
I wrote the question exactly as it reads.

I don't understand what I can really do only given one known value.
 
  • #4
Anyone have any suggestions ?
 
  • #5
Yes, use Gauss! convert units to V/m , and obtain Q as a function of R .
then answer #3 .
skip #2 , unless you want to email your instructor and ask what the sphere radius is.
 
  • #6
Ok, so I converted 180 Volts/in and came out with 7086.61 Volt/meter.

7086.61 = 8.99x109 ( q / r2) ?

Do I know what q is ?

I think I've confused myself.
 
  • #7
Gauss' Law situations typically yield a _surface_charge_density_ "sigma".
When your teacher tells you the radius, then you tell them |Q| .
When your teacher tells you the DIRECTION of the Electric field,
you can tell them the SIGN of the charge ... right?
If you know Q , do you know how to get the number of excess electrons?

Is there a diagram with this question, showing the size R and direction of E ?
 
  • #8
No, there is no diagram along with it. It reads exactly how I wrote the original question.
 
  • #9
It is INCOMPLETE, then . Quit wasting time on it.

do you know the answer to part#3 ? if so, move on.
 
  • #10
The problem is, that it is a take home test and he said we were allowed to use any external source we want.

I emailed another person in my class to see if he is having the same problem.
 

Related to Solving Spherical Conductor Homework: E, q, r2, C, ε₀

1. What is the formula for calculating the electric field (E) of a spherical conductor?

The formula for calculating the electric field of a spherical conductor is E = q / (4πε₀r²), where q is the charge of the conductor, r is the radius of the conductor, and ε₀ is the permittivity of free space.

2. How do you calculate the total charge (q) of a spherical conductor?

The total charge of a spherical conductor can be calculated by using the formula q = 4πε₀r²E, where r is the radius of the conductor and E is the electric field at the surface of the conductor.

3. What is the significance of the radius squared (r²) in the equations for a spherical conductor?

The radius squared (r²) in the equations for a spherical conductor represents the surface area of the conductor. This is important because the electric field and charge of a conductor are directly proportional to its surface area, meaning that a larger radius will result in a stronger electric field and a larger charge.

4. How do you calculate the capacitance (C) of a spherical conductor?

The capacitance of a spherical conductor can be calculated by using the formula C = 4πε₀r, where r is the radius of the conductor and ε₀ is the permittivity of free space. This formula can also be written as C = q / V, where q is the charge of the conductor and V is the potential difference between the conductor and its surroundings.

5. What is the role of permittivity (ε₀) in the equations for a spherical conductor?

Permittivity (ε₀) is a constant that represents the ability of a material to store electric charge. In the equations for a spherical conductor, permittivity is used to calculate the strength of the electric field and the capacitance of the conductor. It is an important factor in determining the behavior of a spherical conductor in an electric field.

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