- Forums
- Homework Help
- Introductory Physics Homework Help

- Thread starterlangenase
- Start date
- Tags
- ConductingConducting sphereSphere

In summary: Then you have E. But what is the electric field?Voltage due to a point charge = kQ/r. The electric field is the intensity of the electric field at a given point. It's just the product of the potential energy and the electric field itself.

- #1

langenase

- 5

- 0

## Homework Statement

A 32 cm diameter conducting sphere is charged to 500 V relative to V=0 and r = infinity.

(a) What is the surface charge density sigma?

(b) At what distance will the potential due to the sphere be only 10 V?

## Homework Equations

## The Attempt at a Solution

I know that to find the surface charge density I have to find charge/area. (Sigma = q/A) Finding the area just means finding the surface area of the sphere right? 4(pi)((0.16m)^2)

Is the charge 500 V? It seems like charge should be expressed in coulombs. I'm not really sure how to get the charge of the sphere.

Then to find the distance at which the potential is only 10 V

Do I use V= (1/4(pi)(eo)* sigma/r? where r is that distance and V = 10V?

Physics news on Phys.org

- #2

learningphysics

Homework Helper

- 4,098

- 7

Hint: The field outside the sphere is the same as the field due to a point charge where the center of the sphere is... where the point charge is the total charge on the sphere...

Using this, you can find the total charge on the sphere... then the charge density.

- #3

Dick

Science Advisor

Homework Helper

- 26,258

- 623

To get the charge, remember Gauss' law?? The rest of what you're doing seems about right. Remember you are given the sphere diameter in cm not m.

- #4

langenase

- 5

- 0

The field outside the sphere is given by the equation E=k(Q/r^2)

I solved for Q and got units in C*m. Was it wrong for me to set E=500V?

- #5

learningphysics

- 4,098

- 7

langenase said:

The field outside the sphere is given by the equation E=k(Q/r^2)

Cool. That's right.

I solved for Q and got units in C*m. Was it wrong for me to set E=500V?

Yeah, that's wrong... what's the voltage due to a point charge... taking the voltage at r = infinity to be 0.

- #6

Dick

Science Advisor

Homework Helper

- 26,258

- 623

Once you've straightened all of this out, you may want to think about the Gauss law approach. Charge contained in a sphere at the surface is (500V)*A*e0 where A is the area. To convert that to surface charge just divide out the A. Using the point charge will (of course) give you the same thing. It's just a little bit longer way around.

- #7

learningphysics

Homework Helper

- 4,098

- 7

Dick said:

Once you've straightened all of this out, you may want to think about the Gauss law approach. Charge contained in a sphere at the surface is (500V)*A*e0 where A is the area. To convert that to surface charge just divide out the A. Using the point charge will (of course) give you the same thing. It's just a little bit longer way around.

But the charge is E*A*e0 according to Gauss law... I'm probably missing something...

- #8

Dick

Science Advisor

Homework Helper

- 26,258

- 623

learningphysics said:

But the charge is E*A*e0 according to Gauss law... I'm probably missing something...

Sorry. I meant surface charge density, which is what the OP is asked for.

- #9

learningphysics

Homework Helper

- 4,098

- 7

Dick said:

Sorry. I meant surface charge density, which is what the OP is asked for.

But the 500V is voltage not electric field...

- #10

Dick

Science Advisor

Homework Helper

- 26,258

- 623

learningphysics said:

But the 500V is voltage not electric field...

Right, J/C not N/C. Please ignore me, I'm apparently having a fuzzy day. Thanks.

- #11

learningphysics

Homework Helper

- 4,098

- 7

Dick said:

Right, J/C not N/C. Please ignore me, I'm apparently having a fuzzy day. Thanks.

No prob. Your fuzzy days are quite rare from what I've seen of your posts.

- #12

langenase

- 5

- 0

I'm a little confused about what to set E equal to. The voltage due to a point charge is when r is infinity is 0. But how does that relate to the electric field?

- #13

learningphysics

Homework Helper

- 4,098

- 7

langenase said:

I'm a little confused about what to set E equal to. The voltage due to a point charge is when r is infinity is 0. But how does that relate to the electric field?

Voltage due to a point charge = kQ/r.

You have voltage... you have r, you can solve for Q.

## FAQ: Calculating Surface Charge Density and Distance in a Conducting Sphere Problem

## 1. What is the conducting sphere problem?

The conducting sphere problem is a mathematical problem in electrostatics that involves finding the electric potential and electric field inside and outside of a conductive sphere. It is a common problem in physics and engineering.

## 2. What is the solution to the conducting sphere problem?

The solution to the conducting sphere problem involves using the boundary conditions and the Laplace or Poisson equations to determine the electric potential and electric field at different points inside and outside of the sphere. The exact solution depends on the specific boundary conditions and assumptions made.

## 3. What are the assumptions made in solving the conducting sphere problem?

Some common assumptions made in solving the conducting sphere problem include assuming the sphere is a perfect conductor, the problem is in a vacuum, and there are no external charges or fields present. These assumptions simplify the problem and allow for a more straightforward solution.

## 4. How does the radius of the conducting sphere affect the solution to the problem?

The radius of the conducting sphere affects the solution to the problem because it determines the size and shape of the electric field. A larger radius will result in a larger electric field and a smaller radius will result in a smaller electric field.

## 5. What are some real-life applications of the conducting sphere problem?

The conducting sphere problem has various real-life applications, including in the design of electronic devices and in understanding the behavior of charged particles in a conductor. It is also used in the study of planetary and celestial bodies, as well as in understanding the behavior of lightning strikes.

### Similar threads

Induced charge on a conducting sphere sliced by a plane

- Replies
- 2

- Views
- 588

Charge on a grounded conducting sphere in a uniform electric field, after ungrounding and movement

- Replies
- 1

- Views
- 729

Charge of 2 conducting spheres separated by a distance

- Replies
- 1

- Views
- 1K

Electric field external to Conducting Hollow Sphere with charge inside

- Replies
- 23

- Views
- 1K

Electric field direction on a grounded conducting sphere

- Replies
- 4

- Views
- 1K

Calculating the surface charge of a sphere and a conducting shell

- Replies
- 5

- Views
- 3K

Kinetic and potential energy of a particle attracted by charged sphere

- Replies
- 6

- Views
- 1K

Attractive force between a charge q and neutral conducting sphere

- Replies
- 5

- Views
- 4K

I Thought I Understood Gauss' Law (Sheets)

- Replies
- 26

- Views
- 1K

Electric field at the center of a sphere

- Replies
- 6

- Views
- 2K

- Forums
- Homework Help
- Introductory Physics Homework Help