A sphere of radius a carries a volume charge density p=p (r/a)^2. Find the electric field inside and outside the sphere.
I don't know what equation to use. Do I use electirc field due to point charge?
(4/3) p * pi r^5 /a^2
Inside the sphere, any imaginary volume you think of will have no charge inside it. Therefore no Electric Field is coming out of it. Thus there is no electric field within the sphere.
The instant you get outside the sphere, the charge looks like it is all at the center, and it is the usual 9*10^9 q/r^2 kind of field.
so the electic field outside the sphere is 9*10^9 p (r/a)^2/a^2?
Whoops, sorry, was thinking charge on surface of sphere.
If uniformly distributed throughout the volume, then my answer is the same outside the sphere
However inside the sphere you will surround chagres the will look as if they are at the center
The amount you surround will be the volume at radius r times the charge density.
V = (4/3) pi r^3
q =V (charge density)= (4/3) pi r^3 (p) (r^2/a^2)
which is
(4/2) p *pi ^5/a^2
that times is q
E = 9*10^9 q/r^2
= 9 * 10^9 (4/3)p* pi r^3/a^2
when you reach r = a
you get
E = 9*10^9 (4/3) p* pi r^3 /a^2
which is of course
9*10^9 ( q total)/r^2 from then on
could i get some help damon
The idea is that inside the sphere, the only charge that counts is what is inside the radius you are at. That looks like it is at the center and is the charge density times the volume inside the radius you are at.
Outside the sphere, the entire charge, the density times the entire volume of the sphere (4/3) pr a^3 * density is inside your radius and no longer increases/
The answers Dan H gave you looked reasonable to me Rory.
yep i noticed that =P
To find the electric field inside and outside the sphere with the given volume charge density, you'll need to use Gauss's Law instead of the equation for electric field due to a point charge.
Gauss's Law states that the electric flux through a closed surface is equal to the total charge enclosed by that surface divided by the electric constant (ε₀). By using a Gaussian surface, which is a hypothetical surface with a symmetric shape around the charge distribution, you can simplify the calculation of the electric field.
Let's start by finding the electric field inside the sphere:
1. Choose a Gaussian surface that is a concentric sphere with a radius 'r' (where r < a so that it's inside the larger sphere).
2. The electric field inside the Gaussian surface will have the same magnitude and direction everywhere due to the spherical symmetry of the charge distribution.
3. Applying Gauss's Law, the electric flux Φ through the Gaussian surface is given by Φ = E × A, where A is the area of the Gaussian surface and E is the magnitude of the electric field.
4. Since the electric field is constant over the Gaussian surface, Φ = E × 4πr².
5. The charge enclosed within the Gaussian surface is given by q_enclosed = ∫p dV, where p is the volume charge density and dV is an infinitesimal volume element.
6. Since the charge density is given as p = p × (r/a)², the total charge enclosed is q_enclosed = p × V_enclosed. Here, V_enclosed is the volume of the Gaussian surface.
7. Express the volume in terms of 'r' using V_enclosed = (4/3)πr³.
8. Substitute q_enclosed = p × V_enclosed into Gauss's Law: E × 4πr² = (p × V_enclosed) / ε₀.
9. Rearrange the equation to solve for E: E = (p × r) / (3ε₀).
Now, let's find the electric field outside the sphere:
1. Choose a Gaussian surface that is a concentric sphere with a radius 'r' (where r > a to be outside the larger sphere).
2. Similar to the previous case, the electric field will be constant in magnitude and direction due to symmetry.
3. Apply Gauss's Law again, Φ = E × A, where A is the area of the Gaussian surface.
4. The charge enclosed within the Gaussian surface in this case is the total charge of the sphere, so q_enclosed = (4/3)πa³p.
5. Express the area in terms of 'r' using A = 4πr².
6. Substitute q_enclosed and A into Gauss's Law: E × 4πr² = ((4/3)πa³p) / ε₀.
7. Solve for E: E = (a³p) / (3ε₀r²).
So, the electric field inside the sphere is E = (p × r) / (3ε₀), and the electric field outside the sphere is E = (a³p) / (3ε₀r²).