Electric field and potential outside and inside nonuniform spherical shell

In summary, using Gauss' equation for the electric field and integrating the charge density over the enclosed volume, we can find the electric field inside and outside a hollow spherical shell with charge density proportional to the distance from the origin. Inside the shell, the electric field is 0, while for a<r<b, it is given by (k(r^4-a^4)/(4e*(r^2)). Outside the shell, it is given by (k(b^4-a^4))/(4e*(r^2)) for all r such that a<b<= r.
  • #1
krickette
3
0

Homework Statement


Find the electric field and potential inside and outside a hollow spherical shell (a [tex]\leq[/tex] r [tex]\leq[/tex] b) which carries a charge density proportional to the distance from the origin in the region, [tex]\rho[/tex]=kr, for some constant k.

Homework Equations


The Attempt at a Solution


I think I've got the electric field inside the shell figured out, but I could be way off base. I have no idea how to do the rest...
help?

also, sorry, for some reason the <or= sign between a and r isn't working
 
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  • #2
Remember Gauss' equation for the electric field: E=Q/(4pi e*r^2), where Q is the charge enclosed, pi=3.14159..., e* is a constant (the permittivity of free space), and r is the distance from the "center" of the charge distribution (analogous to center of mass). Inside the spherical shell (r<a<b), E=0 (no charge is enclosed). For a<r<b, the charge enclosed is obtained by integrating the charge density over the volume enclosed. In this case, the charge density is kr, and the differential element of volume is 4(pi)r^2 x incremental change in r (dr). This yields an enclosed charge of (pi)k(r^4-a^4), and substituting into Gauss' equation yields E=(k(r^4-a^4)/(4e*(r^2)). Outside the spherical shell, the enclosed charge is given by Q=(pi)k(b^4-a^4), and this yields E=(k(b^4-a^4))/(4e*(r^2)) for all r such that a<b<= r.
 

1. What is the difference between electric field and potential?

Electric field is a vector quantity that describes the strength and direction of the force exerted on a charged particle. Potential, on the other hand, is a scalar quantity that represents the amount of work needed to move a charged particle from one point to another in an electric field.

2. How is the electric field outside and inside a nonuniform spherical shell calculated?

Outside the spherical shell, the electric field is calculated using Coulomb's law, which states that the electric field is directly proportional to the magnitude of the charge and inversely proportional to the square of the distance between the charge and the point of observation. Inside the shell, the electric field is zero as there is no net charge enclosed within the Gaussian surface.

3. What determines the magnitude and direction of the electric field outside and inside a nonuniform spherical shell?

The magnitude and direction of the electric field are determined by the distribution of charge on the spherical shell. The electric field lines point away from positive charges and towards negative charges, and the strength of the field is stronger where the charge is more concentrated.

4. How does the electric potential change as you move from the outside to the inside of a nonuniform spherical shell?

The electric potential decreases as you move from the outside to the inside of a nonuniform spherical shell. This is because the electric potential is directly proportional to the electric field, and since the electric field is zero inside the shell, the potential also becomes zero.

5. How can the electric potential be used to determine the direction of the electric field outside and inside a nonuniform spherical shell?

The electric potential can be used to determine the direction of the electric field using the concept of equipotential surfaces. The electric field is always perpendicular to the equipotential surfaces, and the direction of the electric field points towards decreasing potential, i.e. from positive to negative potential.

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