Confirmation of functions for electric potential and field please

In summary: Hint: First find the charge on each surface of the shell, then use the formula for the potential due to a point charge. Remember, the potential at infinity is zero.For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.This is correct.For a < r, the potential difference follows a curve from 30 μC /(4ΠΕ0 R) to 0 as r nears infinity.This is not correct. The potential does not go to zero as r goes to infinity. It goes to a finite value. What does it go to? Hint: Consider the
  • #1
cyberstudent
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A metal sphere of radius R with a charge q is surrounded by a concentric metal sphere as shown in Figure P.23. The outer surface of the spherical shell has a charge of + 30.0 μC and the inner surface of the shell has a charge of +25.0 μC.
b is the radius of the inner surface of the shell and a is the radius of the outer surface of the shell.

(a) Find q

q = -25.0 μC, since a conductor has no electric field. Only the charge on the inner surface of the outer shell affects the metal sphere inside.

(b) Sketch qualitative graphs of (i) the radial electric field component Er and (ii) the electric potential V as functions of r.

This is trickier. Let’s start with V(R)

For r < R (inside sphere), V(r) = -25.0 μC/(k R)
This is because the interior of a conductor has the same potential everywhere. The potential is equal to the potential on the surface.

For R < r < b, the potential goes from -25.0 μC/(k R) to + -25.0 μC/(k R). Right? Very unsure about this.

For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.
Hmmm?

For a < r, the potential difference follows a curve from 30 μC/(k R) to 0 as r nears infinity.

Right?


OK. I know I messed up but I’m still going to try E(R)

For r < R (inside sphere), E(r) = 0
The electric field within a conductor is always 0.

For R < r < b, the potential goes from -25.0 μC/(k R^2) to + -25.0 μC/(k R^2). Right? Very unsure about this.

For b < r < a, E(r) = 0
The electric field within a conductor is always 0.

For a < r, the potential difference follows a curve from 30 μC/(k R^2) to 0 as r nears infinity.

I uploaded my graphs, but I’m not sure if people are able to see my attachments.
 

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  • #2
It looks like you have some R^2 where you should have just R. The contribution to the potential inside any uniform spherical surface distribution is constant, but not zero (unless the charge is zero). There will be a constant contribution to the potential for r<a from the charge at a, etc. Looks like you have left out this contribution..
 
  • #3
Oups. Original answer full of typos.

I didn't sleep trying to do this assignment. :frown: Here is the corrected version without the typos.

A metal sphere of radius R with a charge q is surrounded by a concentric metal sphere as shown in Figure P.23. The outer surface of the spherical shell has a charge of + 30.0 μC and the inner surface of the shell has a charge of +25.0 μC.
b is the radius of the inner surface of the shell and a is the radius of the outer surface of the shell.

(a) Find q

q = -25.0 μC
Since there is no electric field inside a conductor, only the charge on the inner surface of the outer shell affects the metal sphere inside.

(b) Sketch qualitative graphs of (i) the radial electric field component Er and (ii) the electric potential V as functions of r.

Let’s start with V(R)

For r < R (inside sphere), V(r) = -25.0 μC/(4ΠΕ0 R)
The potential is constant and equal to the potential on the surface of the sphere.

For R < r < b, the potential goes from -25.0 μC/(4ΠΕ0 R) to + 25.0 μC/(4ΠΕ0 R).

For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.

For a < r, the potential difference follows a curve from 30 μC/(4ΠΕ0 R) to 0 as r nears infinity.

---------------

E(R)

For r < R (inside sphere), E(r) = 0
The electric field within a conductor is always 0.

For R < r < b, the electric field goes from -25.0 μC/(4ΠΕ0 R^2) to + 25.0 μC/(4ΠΕ0 R^2).

For b < r < a, E(r) = 0
The electric field within a conductor is always 0.

For a < r, the electric field follows a curve from 30 μC/(4ΠΕ0 R^2) to 0 as r nears infinity.

It is easier to follow on the attached graph but I'm still not sure if people can see my attachments.
 

Attachments

  • graphs.gif
    graphs.gif
    11.2 KB · Views: 517
  • #4
cyberstudent said:
A metal sphere of radius R with a charge q is surrounded by a concentric metal sphere as shown in Figure P.23. The outer surface of the spherical shell has a charge of + 30.0 μC and the inner surface of the shell has a charge of +25.0 μC.
b is the radius of the inner surface of the shell and a is the radius of the outer surface of the shell.

(a) Find q

q = -25.0 μC
Since there is no electric field inside a conductor, only the charge on the inner surface of the outer shell affects the metal sphere inside.
Correct q. I suggest a different wording. It sounds like you are saying that the charge on the inner surface of the shell is causing the charge to appear on the inner sphere. In fact, it is the other way around. A charge placed on the inner sphere is what causes the charge to distribute on the inner surface of the conducting shell. Without the charge on the inner sphere, there is no way you can get any charge to stay on the inner surface of the shell.
cyberstudent said:
(b) Sketch qualitative graphs of (i) the radial electric field component Er and (ii) the electric potential V as functions of r.

Let’s start with V(R)

For r < R (inside sphere), V(r) = -25.0 μC/(4ΠΕ0 R)
The potential is constant and equal to the potential on the surface of the sphere.
This is not correct. The potential is constant, but not the value you state.
cyberstudent said:
For R < r < b, the potential goes from -25.0 μC/(4ΠΕ0 R) to + 25.0 μC/(4ΠΕ0 R).
This is not correct. The potential does change in this region, but not the way you say. What is the equation that describes the changing potential?
cyberstudent said:
For b < r < a, the potential difference is constant. V(r) = 30.0 μC - 25.0 μC = 5 μC.
The potential is constant in this region, but charge is not potential. Reserve the phrase "potential difference" for talking about the difference in potential between two points. By convention, we choose to define the potential due to any configuration of charges to be zero at infinity, and talk about the absolute potential at other points. It is correct to say the potential difference between any two points in this region is zero, or to say the potential in this region is constant.
cyberstudent said:
For a < r, the potential difference follows a curve from 30 μC/(4ΠΕ0 R) to 0 as r nears infinity.
This is not correct. You have used to wrong radius. What is the equation of the curve?

I think you understand that for one spherical surface with a uniform distribution of charge the potential is inversely proportional to the distance from the center at points outside the sphere and constant inside the sphere. I suggest you start in the r > a region and work inwards, keeping in mind that each surface makes a constant contribution to the potential inside of that surface. Any variation in potential in a region is the result of the net charge closer to the center of the spheres.
cyberstudent said:
E(R)

For r < R (inside sphere), E(r) = 0
The electric field within a conductor is always 0.
This is correct.
cyberstudent said:
For R < r < b, the electric field goes from -25.0 μC/(4ΠΕ0 R^2) to + 25.0 μC/(4ΠΕ0 R^2).
This is not correct. The outer shell contributes nothing to the electric field inside its cavity. What is the equation that describes the electric field in this region?
cyberstudent said:
For b < r < a, E(r) = 0
The electric field within a conductor is always 0.
This is correct.
cyberstudent said:
For a < r, the electric field follows a curve from 30 μC/(4ΠΕ0 R^2) to 0 as r nears infinity.
This is not correct. You have used the wrong radius. What is the equation that describes the electric field in this region?
cyberstudent said:
It is easier to follow on the attached graph but I'm still not sure if people can see my attachments.
I still cannot see them, but I know what they are supposed to look like.
 

FAQ: Confirmation of functions for electric potential and field please

What is electric potential and how is it different from electric field?

Electric potential is the amount of work required to move a unit positive charge from one point to another in an electric field. It is a scalar quantity, meaning it has only magnitude and no direction. Electric field, on the other hand, is a vector quantity that describes the force experienced by a unit charge at a given point in an electric field.

How is electric potential and field measured?

Electric potential is measured in units of volts (V) using a voltmeter. Electric field is measured in units of volts per meter (V/m) using an electrostatic voltmeter or a device called an electroscope.

What is the relationship between electric potential and field?

Electric potential is directly related to electric field through the equation V = Ed, where V is electric potential, E is electric field, and d is the distance between the two points. This means that the electric field at a point is equal to the change in electric potential per unit distance.

How do electric potential and field affect charged particles?

Electric potential and field can affect charged particles in a variety of ways. For example, they can cause a charged particle to experience a force, resulting in its movement. They can also affect the speed and direction of charged particles in an electric field.

What are some real-world applications of electric potential and field?

Electric potential and field have numerous practical applications, including powering electrical devices, generating electricity, and controlling the movement of charged particles in technologies such as particle accelerators. They are also important in understanding the behavior of lightning, the functioning of batteries, and the operation of electronic devices.

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