Gauss's Law and charge distribution

[KillerZ]

Homework Statement

A spherically symmetric charge distribution produces the electric field $$E = (5000r^2) N/C$$, where r is in m.

a) What is the electric field strength at r = 20cm?

b) What is the electric flux through a 40cm diameter spherical surface that is concentric with the charge distribution?

c) How much charge is inside this 40cm diameter spherical surface?

Homework Equations

$$E = (5000r^2) N/C$$

The Attempt at a Solution

I am not sure if I did this right?

a)

$$E = (5000r^2) N/C$$

$$E = (5000(0.2m)^2) N/C$$

= $$200 Nm^2/C$$

b)

40cm = diameter
20cm = radius

$$E = (5000(0.2m)^2) N/C$$

= $$200 Nm^2/C$$

c)

$$\phi_{e}=\frac{q}{\epsilon_0}$$

$$q=(\phi_{e})(\epsilon_0)$$

$$q=(200 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)$$

$$= 1x10^{-9} C$$

 

[Dick]

You are stumbling on the units. They should have written E=5000*(r/m)^2*N/C, in my opinion. I think that's what they really mean by "where r is in m". Now if you do a) you get E=5000*(0.2m/m)^2*N/Q. Now you have correct E field units of N/C. For the flux in b) you need to integrate that field over the surface of the sphere. Now you pick up the extra m^2 in the units. But you have to multiply the E field from a) by the area of the sphere.

 

[KillerZ]

Ok, I redid it:

a)

$$E = (5000r^2) N/C$$

$$E = (5000(0.2m/m)^2) N/C$$

= $$200 N/C$$

b)

40cm = diameter
20cm = radius

$$\phi_{e}=\oint EdA = EA_{sphere}$$

= $$(200 N/C)(4\pi(0.2m)^2)$$

=$$101 Nm^2/C$$

c)

$$\phi_{e}=\frac{q}{\epsilon_0}$$

$$q=(\phi_{e})(\epsilon_0)$$

$$q=(101 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)$$

$$= 8.9x10^{-10} C$$

 

[Dick]

That looks much better.