Calculating Electric Field: Gauss's Law with Uniformly Distributed Charge

In summary, the problem involves a charge of 9 pC uniformly distributed between two concentric spherical surfaces, with radii of 1.8 cm and 3.8 cm. The task is to find the magnitude of the electric field at a point 2.9 cm from the center of the surfaces, using the equations for electric flux and electric field. The solution involves solving for the theoretical inner charge of the inner sphere and then using the formula for electric field at a point created by a charge to find the electric field at the given point. The Gauss' Law can also be used to solve this problem.
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Soulis
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Homework Statement


A charge of 9 pC is uniformly distributed
throughout the volume between concentric
spherical surfaces having radii of 1.8 cm and
3.8 cm.

Let: Ke = 8.98755 × 109 N · m2/C2 .
What is the magnitude of the electric field
2.9 cm from the center of the surfaces?
Answer in units of N/C

Homework Equations


Electric Flux: (Ie) = E*A = q/epsilon
Electric Field: E = Ke*q/r^2

where A = Surface Area of gaussian sphere, and epsilon is a constant = 8.8542e-12

The Attempt at a Solution


My approach to this problem was assuming their was an inner charge at the center, which I labeled q. To start I used the formula for net flux Ie (electric flux) = E*A = q/epsilon. I neglected the outer sphere completely and used the principle that E (the electric field) is the electric field just outside the conductor and let that equal the charge given in the problem, and epsilon is a constant equal to 8.8542e-12, and A is the surface area of the inner sphere of radius .018m.

E = 9 pC (9e-12)
A = 4pi*r^2 = .004071504079
q = ?
epsilon = 8.8542e-12

Solved for q (my theoretical inner charge)

I used this equation to find my theoretical inner charge of the inner sphere (q) and then applied the generic formula for an electric field at a point P created by a charge q, E = Ke*q/r^2 to solve for the Electric field at P(a distance r from q).

Ke = 8.98755e+9
q = 3.244492027e-25 (answer from first part)
r = .029m

Using this approach I came out with E=3.467304913e-12, which needless to say, was incorrect. I feel I may have overcomplicated this problem and am approaching it incorrectly. Any help would be greatly appreciated, thanks.
 
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Can you formulate Gauss' Law?

ehild
 

Related to Calculating Electric Field: Gauss's Law with Uniformly Distributed Charge

What is Gauss's Law?

Gauss's Law is a fundamental law in physics that relates the electric field to the distribution of electric charges. It states that the electric flux through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space. This law was developed by the German mathematician and physicist Carl Friedrich Gauss in the early 19th century.

What is the significance of Gauss's Law?

Gauss's Law is significant because it provides a mathematical relationship between electric charges and the resulting electric fields. It is one of the four Maxwell's equations that form the foundation of classical electromagnetism. This law also helps in calculating the electric field in a variety of situations, such as for a point charge, a charged sphere, or a charged cylinder.

How do we apply Gauss's Law?

To apply Gauss's Law, we first need to choose a closed surface that encloses the charge distribution we want to analyze. Then, we calculate the electric flux through that surface by taking the dot product of the electric field and the surface area. Finally, we equate the electric flux to the net charge enclosed by the surface divided by the permittivity of free space to solve for the electric field.

What are the limitations of Gauss's Law?

Gauss's Law is limited to situations with symmetrical charge distributions, such as a point charge, a charged sphere, or a charged cylinder. It cannot be applied to more complex charge distributions, such as non-uniformly charged objects or systems with moving charges. Additionally, Gauss's Law assumes a vacuum or a uniform dielectric medium, so it may not be accurate in materials with varying permittivity.

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

Gauss's Law and Coulomb's Law are related through the concept of electric flux. Coulomb's Law states that the magnitude of the electric force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Gauss's Law, on the other hand, relates the electric flux to the enclosed charge. By using Gauss's Law, we can derive Coulomb's Law for a point charge.

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