# Electric field sphere with inner cavity

1. Oct 13, 2009

### jackxxny

1. The problem statement, all variables and given/known data
I have :
Spherical conductor has a inner cubic cavity of side = b;
inside the cubic cavity we have a charge = y;
the surface of the sphere has a charge density = z;
I need to calculate the electric field at some point g, where g>r;

2. Relevant equations

3. The attempt at a solution

E=(1/4*pi*eo)*∫ p d(tao)

= (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi)

will that work?

2. Oct 13, 2009

### kuruman

You did not define point "r", so it is not clear where "g" is. If g > x, then you can use Gauss' Law. Th shape of the cavity does not affect the electric field outside the spherical conductor.

3. Oct 13, 2009

### jackxxny

g is a point outside the sphere.

so do i use just

E= y/4*pi*eo*x^2

or

= (1/4*pi*eo)*∫chargedensity(z) *radius(x)^2 * sin(theta) d(x) d(theta) d (phi)

???

my question is how o i use the charge density?

4. Oct 14, 2009

### kuruman

Construct a Gaussian surface of radius g. Use Gauss' Law to say that the total electric flux through the surface of the sphere is equal to the total enclosed charge divided by ε0. You already know that there is charge y inside the cavity. You can then use the charge density to find the additional charge on the conductor. Add y to that and you have the total enclosed charge.

5. Oct 14, 2009

### jackxxny

what about the electric field....i am so lost...

6. Oct 14, 2009

### kuruman

WHat about it? What does Gauss' Law? How do you use it when you have a spherical distribution?

7. Oct 14, 2009

### jackxxny

gauss law for a sphere is

E= pR3/(3eor2)

that's all i have to do?

8. Oct 14, 2009

### kuruman

That is not Gauss' Law for a sphere. That is the electric field outside a uniformly charged sphere of volume charge density ρ.

Gauss' Law for a uniform spherical distribution says that the field outside the distribution is given by

E (4πr2)=qenclosed0

where qenclosed is the charge enclosed by a Gaussian surface of radius r. E in this case is the electric field at radius r. Reread my second posting and do what I suggested to finish this problem.

9. Oct 14, 2009

### jackxxny

i think i kinda got it

ε0*E (4πr2)=qenclosed

i can then substitute

qenclose= integral rho d tau

then take the derivative on both sides

right?

10. Oct 14, 2009

### kuruman

What does taking the derivative do for you? You need to find the electric field. Find how much charge is enclosed, then solve for the electric field.

11. Oct 14, 2009

### jackxxny

E =integral rho d tau/((4πr2)*ε0)

sorry i simplify for E

how do i do the integral of rho d tau???

12. Oct 14, 2009

### kuruman

You can find the enclosed charge without doing the integral. Just read the statement of the problem very carefully.

13. Oct 14, 2009

### jackxxny

the problem gave a number for charge density

so should i pull the number in front of the integral and do

$$\int between 0 and x of 4 \ pi r^2 da$$

??

14. Oct 14, 2009

### kuruman

If I can read your mathematical expression correctly, yes integrate the charge density over the area. This is a surface integral, not a volume integral.