How Can A Be Calculated for Uniform Electric Field in a Spherical Shell?

[paralian]

Homework Statement

(Note: Don't worry about significant digits. I just want to be able to do the question and will worry about significant digits on the exam.)

In the figure below, a non-conducting spherical shell of inner radius a=2.00cm and outer radius b=2.40cm has (within its thickness) a positive volume density $$\rho$$=A/r, where A is a constant and r is the distance from the centre of the shell. In addition, a small ball of charge q=45.0 fC is located at the centre.

What value should A have if the electric field in the shell (a$$\leq$$r$$\leq$$b) is to be uniform?

(See attached file for the picture. sorry it isn't very good. That's a "q" beside the +ve charge in the middle.)

Homework Equations

Charge density : $$\rho= \frac{dq}{dV}$$

Volume : $$V = 4\pi r^3$$

Electric field : $$E = \frac{1}{4\pi \epsilon} \int \frac{dq}{r^2}$$

The Attempt at a Solution

E at a distance r anywhere within [a, b] is constant.
E is caused by the point charge and distribution of charges in the volume.

$$E_{net}$$ is constant, therefore

$$E_{charge}=-E_{volume}$$

$$\frac{1}{4\pi \epsilon} \int \frac{dq_{charge}}{r_{charge}^2} = - \frac{1}{4\pi \epsilon} \int \frac{dq_{volume}}{r_{volume}^2}$$

the E for a point charge doesn't need to have an integral, because it works out that way.

$$\frac{q_{charge}}{r^2} = - \int \frac{dq_{volume}}{r_{volume}^2}$$

Charge density : $$\rho= \frac{A}{r} = \frac{dq}{dV}$$

$$dq = \frac{A}{r} dV$$

$$\frac{q_{charge}}{r^2} = - \int \frac{A dV}{r^3}$$

dV is related to r because $$V = \frac{4}{3} \pi (r^3 - a^3)$$

then $$r^3 = \frac{3V + 4 \pi a^3}{4 \pi}$$

$$\frac{q_{charge}}{r^2} = - \int \frac{4 \pi A dV}{3V + 4 \pi a^3}$$

$$\frac{q_{charge}}{r^2} = - 4 \pi A \int \frac{dV}{3V + 4 \pi a^3}$$

$$\frac{q_{charge}}{r^2} = - \frac{4 \pi A}{3} ln (3V + 4 \pi a^3)$$

I don't know what to do after this, and I don't know what to do. I don't know if I messed up something.

This may be supposed to be done with Gauss' Law, but I couldn't figure out what to do with that.

Gauss' Law : $$\Phi = \oint E da$$ (In this case since E and a are in the same direction, the dot product just became E da)

 

[Nate810]

\Phi = \oint E da = \int_a^b E 4 \pi r^2 dr = \frac{q}{\epsilon} I don't know what to do with this either. Can someone help me?

 

[Moetasim]

The electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

Based on the given information, the value of A can be determined by using Gauss' Law. Since the electric field is constant within the shell, we can use a Gaussian surface in the shape of a sphere with radius r, where a≤r≤b. The electric flux through this surface is equal to the charge enclosed by the surface divided by the permittivity of free space.

Therefore, we can write the following equation:

Φ = EA = qenc/ε0

Where qenc is the enclosed charge within the Gaussian surface. In this case, the only charge enclosed is the small ball of charge q at the center.

qenc = q = 45.0 fC = 45.0 x 10^-15 C

Substituting this into the equation, we get:

Φ = EA = (45.0 x 10^-15 C)/ε0

Now, we know that the electric field E is constant within the shell, so we can write:

E = qenc/(4πε0r^2)

Substituting this into the previous equation, we get:

Φ = (qenc/(4πε0r^2))A = (45.0 x 10^-15 C)/ε0

Solving for A, we get:

A = (45.0 x 10^-15 C)4πε0/(ε0r^2) = 45.0 x 10^-15 C/(4πε0r^2)

Since r is the distance from the center of the shell, we can write:

r = (a+b)/2 = (2.00cm + 2.40cm)/2 = 2.20cm

Substituting this into the equation for A, we get:

A = 45.0 x 10^-15 C/(4πε0(2.20cm)^2)

Simplifying, we get:

A = 1.45 x 10^-12 C/m^2

Therefore, the value of A should be 1.45 x 10^-12 C/m^2 for the electric field to be uniform within the shell.