Electric field sphere with inner cavity

[jackxxny]

Homework Statement

I have :
Spherical conductor of radius=x;
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;

Homework Equations

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?

 

[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.

 

[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?

 

[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.

 

[jackxxny]

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

 

[kuruman]

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

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

 

[jackxxny]

gauss law for a sphere is


E= pR³/(3e_or²)


that's all i have to do?

 

[kuruman]

gauss law for a sphere is


E= pR³/(3e_or²)


that's all i have to do?

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)=q_enclosed/ε₀

where q_enclosed 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.

 

[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?

 

[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.

 

[jackxxny]

E =integral rho d tau/((4πr2)*ε0)


sorry i simplify for E

how do i do the integral of rho d tau?

 

[kuruman]

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

 

[jackxxny]

the problem gave a number for charge density


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


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


??

 

[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.