Surface Volume and Line charge densities, how to solve problems?

[turtieari]

This question is a perfect example:

A solid sphere 25cm in radius carries 14microC, distributed uniformly throughout it's volume. Find the electric field strength a) 15cm b)25cm and c)50 cm from center.

I know that I need my gaussian surface and I also need
p=q/v where p is the (Qenclosed) in the equation: Qenclosed/epsilon=EA

Could someone help me understand what the charge densities are for. Do I have to subsitute Qenclosed for p or do I just use it as a clue that I need the volume?

Thank you for your time!

Warmest regards,
ARi :"D

 

[Freespader]

I don't think you actually need to use Gauss's law for this one, TBH. Since it's a sphere with uniform ρ, by the Shell Theorem, it acts like all the charge were centered at, well, the center. I'd solve using

E = q/(4πε₀r^2)

If you really want to use a Gaussian surface, since it's symmetrical:

qenc/ε₀ = EA -> E = 14µC/ε₀(4πR^2)

Where R is .15m+whatever the distance is from the sphere. Since the shell theorem comes from Gauss's Law, sort of, I guess they're actually the same answer, though.

Hope this helps.

 

[turtieari]

The shell theorem is new to me. Maybe I don't know with that name. THANK YOU!

 

[Freespader]

The shell theorem says if you have a "shell", infinitely thin and spherical with uniform charge density, the charge acts as if all the charge were at the center. It also works with rings, but not in the 3rd dimeension. The corollary is that a solid sphere, being an infinite number of shells, also acts as if it were at the center.

We learned it in AP physics, but I don't know if it's a 'real' theorem.

 

[sardar]

Hi ARi,

To solve this problem, you will need to use the concept of charge densities. Charge density is defined as the amount of charge per unit volume or per unit area. In this case, the problem is asking for the electric field strength at different distances from the center of a solid sphere with a given radius and charge.

To start, we need to find the charge density of the sphere. Since the charge is distributed uniformly throughout the volume, we can use the formula p=q/v, where p is the charge density, q is the total charge, and v is the volume of the sphere. In this case, the total charge (q) is given as 14 microC and the volume (v) of a sphere is (4/3)πr^3, where r is the radius. Substituting these values, we get p= (14 microC)/[(4/3)π(25 cm)^3] = 1.12 microC/cm^3.

Now, we can use this charge density to find the electric field strength at different distances from the center of the sphere. To do this, we will use Gauss's Law, which states that the electric flux through a closed surface is equal to the charge enclosed divided by the permittivity of free space (ε0). Mathematically, it can be written as Φ=E*A=qenclosed/ε0, where Φ is the electric flux, E is the electric field strength, and A is the area of the closed surface.

In this problem, we will use a spherical Gaussian surface with different radii to find the electric field strength at different distances. For a) 15 cm, the Gaussian surface will have a radius of 15 cm. Since the charge is uniformly distributed, the enclosed charge will be the same as the total charge, which is 14 microC. Substituting these values in the equation, we get E=(14 microC)/(ε0*4π*(15 cm)^2) = 6.22 N/C.

Similarly, for b) 25 cm, the Gaussian surface will have a radius of 25 cm and the enclosed charge will still be 14 microC. Substituting these values, we get E=(14 microC)/(ε0*4π*(25 cm)^2) = 2.80 N/C.

For c) 50 cm, the Gaussian surface will have a radius of