How Do You Calculate the Charge on the Inner Sphere Using Gauss's Law?

In summary, the conversation discusses the use of Gauss's law to calculate the charge on an inner sphere with a radius of 3.000 cm, given a radial electric field of 7.366e+04 N/C at a distance of 3.125 cm. The mistake of using the area of the inner sphere instead of the Gaussian surface is pointed out, and the topic of magnetism is clarified as being irrelevant to the discussion.
  • #1
frogstomp_theory
Two charged concentric metal spheres have radii of 3.000 cm and 3.500 cm. Calculate the charge in nC on the inner sphere if the radial electric field at a radial distance 3.125 cm is 7.366e+04 N/C.

I tried using Gausses law of EA = Q/e0 then after getting
Q I would take the average charge per area. Then I would simply enter into the equation Q=o(greek letter)A the area of the inner sphere, but that's wrong:frown: any suggestions?

I've been having trouble with magnetism so any help is appreciated, thanks.
 
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  • #2
Hi, and welcome to PF.

Originally posted by frogstomp_theory
I tried using Gausses law of EA = Q/e0 then after getting
Q I would take the average charge per area. Then I would simply enter into the equation Q=o(greek letter)A the area of the inner sphere, but that's wrong:frown: any suggestions?

I've highlighted your mistake above. You should be using the area of the Gaussian surface, not the inner sphere.

I've been having trouble with magnetism so any help is appreciated, thanks.

Actually, this is electrostatics and has nothing to do with magnetism.
 
  • #3


Based on the given information, we can use Gauss's law to calculate the charge on the inner sphere. Gauss's law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In this case, the inner sphere acts as the closed surface, with the radial distance of 3.125 cm being the radius of the surface.

First, we need to calculate the electric flux through the surface. The electric field is given as 7.366e+04 N/C, and the area of the inner sphere can be calculated using the formula for the surface area of a sphere (4πr^2). Plugging in the values, we get the electric flux to be 2.89e+07 Nm^2/C.

Next, we can rearrange Gauss's law to solve for the charge enclosed by the inner sphere. This gives us Q = ε0 * electric flux. Plugging in the values, we get the charge on the inner sphere to be 2.89e+04 C. However, the given unit for charge is in nanocoulombs (nC), so we need to convert it. 1 C = 1e+9 nC, so the charge on the inner sphere is 2.89e+13 nC.

In conclusion, the charge on the inner sphere is 2.89e+13 nC. It is important to double check the units and make sure they are consistent throughout the calculation. Hope this helps!
 

FAQ: How Do You Calculate the Charge on the Inner Sphere Using Gauss's Law?

1. What are two concentric charged spheres?

Two concentric charged spheres are two spheres that are centered at the same point and have opposite charges. They are often used as models in electrostatics experiments to demonstrate and study the behavior of electric fields and charges.

2. How are the charges distributed on two concentric charged spheres?

The inner sphere is typically charged with a positive charge, while the outer sphere is charged with an equal but opposite negative charge. This distribution creates a uniform electric field between the two spheres, with the highest field strength near the surface of the spheres.

3. What is the electric potential between two concentric charged spheres?

The electric potential between two concentric charged spheres is inversely proportional to the distance between the spheres and directly proportional to the charges on the spheres. It can be calculated using the equation V = kQ/R, where k is the Coulomb constant, Q is the charge on the spheres, and R is the distance between the spheres.

4. How does the distance between two concentric charged spheres affect the electric field strength?

The electric field strength between two concentric charged spheres is inversely proportional to the distance between the spheres. This means that as the distance between the spheres increases, the electric field strength decreases. When the spheres are brought closer together, the electric field strength increases.

5. What are some real-world applications of two concentric charged spheres?

Two concentric charged spheres are commonly used in electrostatic experiments to demonstrate concepts such as electric fields, potential, and capacitance. They are also used in high-voltage equipment, such as Van de Graaff generators, and in the design of capacitors used in electronic circuits.

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