# Homework Help: Charge density

1. Oct 13, 2009

### jackxxny

1. The problem statement, all variables and given/known data
i have:
a nonconducting sphere of radius 25 cm. the electric field points radially away from the center of the sphere, and increase linearly from 0 to 3700V/m at the surface of the sphere.
determine charge density?

2. Relevant equations

i think the E=E0+r*148

3. The attempt at a solution

so i did $$\int E *da = \int p(r) d \tau / \epsilon0$$

then i substitute E

$$da = 4* pi * r^{2}$$

$$d \tau = 4* pi * r^{2} dr$$

then i simplify for p(r)

is that right????

2. Oct 13, 2009

### rl.bhat

Yes. It is right.

3. Oct 13, 2009

### jackxxny

thank you

4. Oct 13, 2009

### jackxxny

i got

p(r) = $$\epsilon 0 *(E_{}o 2/r +3)$$

can that be???

and now the problem is asking for the potential difference between the center and the surface of the sphere

i know that

$$\Delta V = -\int E dr$$

will that do it??

5. Oct 13, 2009

### rl.bhat

p(r) = $$\epsilon 0 *(E_{}o 2/r +3)$$
How did you get this one?
Your formula for dV is correct.

6. Oct 13, 2009

### jackxxny

$$(Eo+r*148)*( 4* pi * r^{2})* \epsilon 0 = \int p(r) (4*pi*r^{2}) dr$$

then i take the derivative on both sides and isolate p(r)

i obtain =

$$\epsilon 0*(E0*2/r +396) = p(r)$$

7. Oct 13, 2009

### jackxxny

i got -15000 for the potential. I consider Eo to be zero is that a good assumption?

8. Oct 13, 2009

### rl.bhat

think the E=E0+r*148
Is it given in the problem?

9. Oct 14, 2009

### jackxxny

no. I thought since the problem said that it is linearly that might work?

10. Oct 14, 2009

### rl.bhat

No. It is not correct.
Electric field is given.
You have found E = ρ*r/εο. Find ρ.
Now dV = - dE*dr To find the potential V on the surface find the integration.

11. Oct 14, 2009

### ehild

..."the electric field points radially away from the center of the sphere, and increase linearly from 0 to 3700V/m at the surface of the sphere."

- this means that E=0 at the centre of the sphere and E=Es=3700 V/m at the surface, and increases linearly inside the sphere from r=0 to r=R, the radius of the sphere.
The enclosed charge in the sphere is
$$Q=4/3 R^3 \pi \rho$$ and by Gauss' Law, $$Q=4R^2 \pi \epsilon_0 E_s$$. Comparing the equations for the total charge, you get the charge density.
As for the potential at the surface of the sphere: remember that we usually choose zero potential at infinity, and outside a homogeneously charged sphere the electric field is the same as that of a point charge....

ehild

12. Oct 14, 2009

### jackxxny

$$4/3 R^3 \pi \rho = 4R^2 \pi \epsilon_0 E_s$$

so i just set it equal to each other and solve for p

$$\rho = 3 \epsilon_0 E_s / R$$

is that it?

13. Oct 14, 2009

### ehild

Yes. Now calculate it with the given data.

ehild

14. Oct 14, 2009

### jackxxny

ok i got the charge density

how do i do the potential

so how do i do this integral

dV = - dE*dr

15. Oct 15, 2009

### ehild

Previously you have written the integral correctly:

$$\Delta V = -\int E dr$$

You want the potential at the surface of the sphere and you know that it is nought at infinity:

$$\Delta V= V(\infty)-V(R) = -\int_R^\infty{ E dr}$$

You certainly have learnt that the electric field outside a homogeneously charged sphere is the same as if its total charge were concentrated in the centre. What is the field around a point charge?

$$E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$$,

is it right? You know Q, the total charge of the sphere.

Now integrate this formula for E.

ehild