Gauss's Law and the electric field

In summary, the conversation is about a problem involving determining charge using Gauss's Law. The given values for a, b, and c are 5.00 cm, 20.0 cm, and 25.0 cm respectively, and the electric field is measured to be 3.70*10^3 N/C at a point 15.5 cm from the center, and 2.50*10^2 N/C at a point 50.0 cm from the center. The person is having trouble solving for the charge on the insulating sphere and is asking for help. They mention using Gauss's Law and a constant for epsilon, but are unsure of their calculations and are having trouble with units.
  • #1
Th3Proj3ct
19
0
I'm stuck on a problem trying to determine charge, and I'm hoping someone can help.

Suppose that a = 5.00 cm, b = 20.0 cm, and c = 25.0 cm. Furthermore, suppose that the electric field at a point 15.5 cm from the center is measured to be 3.70 *103 N/C radially inward while the electric field at a point 50.0 cm from the center is 2.50 *102 N/C radially outward.

http://www.webassign.net/pse/p24-57.gif

From this information, find the following charges. (Include the sign of the charges.)
(a) the charge on the insulating sphere


At first i worked out the whole EA=Q/Epsilon(0); and tried 4pi(r^2)*Epsilon*3.7e3 in order to solve for Q, but it didnt work out; what am I doing wrong?
 
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  • #2
It would help to see more of your work. For example, what values do you get and what were you expecting?

The only thing I can suggest, based on what you've told us, is that you be careful with your units. It looks like you're mixing cgs and mks units, so that could trip you up.
 
  • #3
Well for E, i used the 3.7x10^3 N/C, and A I used (3.1416*.05^2*4). Epsilon is a constant, so I figured it would be easy to solve for Q, but I get -1.03x10^-9 C, which is within 10%-100% off; so I know I'm somewhere in the range of the answer.
 
  • #4
How are you using Gauss's Law in this problem? Doesn't it require that the units all be cgs (cm, g, s) or mks (m, kg, s)? You have both N, which is MKS, and cm, which is cgs.

This might not be your problem - the choice of units in E&M is always tricky - but without seeing what else you've done, it's hard for me to tell.
 

What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the distribution of electric charges to the resulting electric field. It states that the electric flux through any closed surface is equal to the net electric charge enclosed by that surface divided by the permittivity of free space.

How do you calculate the electric field using Gauss's Law?

To calculate the electric field using Gauss's Law, you must first choose a closed surface known as a Gaussian surface. The electric field is then calculated by taking the integral of the electric flux over this surface and dividing it by the surface area. This can be represented by the equation E = Q/ε0A, where E is the electric field, Q is the net charge enclosed, ε0 is the permittivity of free space, and A is the surface area.

What is the significance of Gauss's Law?

Gauss's Law is significant because it provides a powerful tool for calculating the electric field in situations with high symmetry, such as when dealing with point charges or charged spheres. It also allows us to understand the relationship between electric charges and the resulting electric field, aiding in the understanding of electrostatics and the behavior of electrically charged particles.

Can Gauss's Law be used in all situations?

No, Gauss's Law can only be used in situations with high symmetry. This means that the electric field must be constant over the chosen Gaussian surface. In more complex situations, other methods such as Coulomb's Law or integration may need to be used to calculate the electric field.

How is Gauss's Law related to Coulomb's Law?

Gauss's Law and Coulomb's Law are closely related, as they both deal with the relationship between electric charges and the resulting electric field. However, Gauss's Law is a more general and powerful law, as it can be applied to situations with high symmetry. Coulomb's Law is a special case of Gauss's Law, where the electric field is constant over a chosen Gaussian surface. This means that Gauss's Law can be derived from Coulomb's Law, but not vice versa.

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