Determining Charge Density of Nonconducting Sphere

[jackxxny]

Homework Statement

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?

Homework Equations

i think the E=E₀+r*148

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}<br /> <br />

<br /> d \tau = 4* pi * r^{2} dr<br />then i simplify for p(r)is that right?

 

[rl.bhat]

Yes. It is right.

 

[jackxxny]

thank you

 

[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??

 

[rl.bhat]

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

 

[jackxxny]

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


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

i obtain =

<br /> \epsilon 0*(E0*2/r +396) = p(r)<br /> <br />

 

[jackxxny]

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

 

[rl.bhat]

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

 

[jackxxny]

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

 

[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.

 

[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=E_s=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

 

[jackxxny]

<br /> 4/3 R^3 \pi \rho = 4R^2 \pi \epsilon_0 E_s <br />

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


<br /> \rho = 3 \epsilon_0 E_s / R <br />

is that it?

 

[ehild]

Yes. Now calculate it with the given data.

ehild

 

[jackxxny]

ok i got the charge density


how do i do the potential


so how do i do this integral

dV = - dE*dr

 

[ehild]

Previously you have written the integral correctly:


<br /> \Delta V = -\int E dr<br />

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

<br /> \Delta V= V(\infty)-V(R) = -\int_R^\infty{ E dr}<br />

You certainly have learned 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?

<br /> <br /> E=\frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}<br /> <br />,

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

Now integrate this formula for E.

ehild