# Homework Help: Some flux questions

1. Aug 4, 2007

### t_n_p

The problem statement, all variables and given/known data

2. A large uniform sheet of charge on an insulator produces a field of 1000 V/m directed towards the sheet, at a distance 2 mm in air. What is its charge in microcoulomb per square metre? Take care with the sign.

3. A spherical balloon of radius R = 2 m has a uniformly charged surface of total charge +Q. The electric field strength is 3.0 kV/m directed outward at a radial distance 4R from the balloon's centre.
If the balloon is now is inflated so that its new radius is 1.1R, and its charge is doubled, what now is the field at the same radial distance 4R from the balloon's centre, in kV/m?

The attempt at a solution

2. I think this is the correct method. From Gauss's Law I use 2EA = enclosed charge/epsilon0, giving enclosed charge/area = -1.77*10^-8. Converting that to microcoulomb/square metre gives -1.77*10^-14. I think it's right, but not 100%

3. Got no idea!

Edit: I solved the first question.

Last edited: Aug 5, 2007
2. Aug 5, 2007

### chaoseverlasting

2. The field should be EA and not 2EA, you only need to consider one surface and not two.

3. Use coulombs law. And the change in radius doesnt matter, you only need to consider the distance from the center of the sphere.

3. Aug 5, 2007

### t_n_p

Thanks!
For 2, I figured out the answer to be 8.85*10^-15 nC/m^2 using EA rather than 2EA.

For 3, After using coulombs law, what forumla do I use to get the field in kV/m?

4. Aug 6, 2007

### andrevdh

2. Since one is closeby the sheet I think that the intention is using the derived formula for a large sheet

$$E = \frac{\sigma}{2\varepsilon _o}$$

you are correct in making the answer negative since the sheet needs to be negatively charged (a positive test charge will be attracted towards the sheet)

5. Aug 6, 2007

### andrevdh

3. Here you need to reason what happens to the flux. If the charge on the balloon stayed the same the flux would not have changed even if the baloon is inflated - according to Gauss' law. The flux only changes (for the same gaussian surface) if the enclosed charge is altered. Since the field is required at the same distance the flux will not be influenced by the area component of the summation, only the electric field in the summation (integration) changes the flux. This means that doubling the charge would result in the electric field being ...