How to Find Force on an Electron Using Gauss' Law?

[ichivictus]

Homework Statement

The charge density of a spherically symmetric and uniform distribution of charge is ρ = 2*10^-3 C/m^3. An electron is released one centimeter from the center. Find the magnitude and direction of the force onto the electron.

Homework Equations

charge density = Q/V = Q/[(4/3)pi*r^3]

EA = Q(V_gs/V_s)
where E is the electric field, A is the area of sphere, Q is the charge of field, and the ratio is the volume of gaussian surface divided by the volume of entire surface.

Elementary charge constant for an electron = -1.602 * 10^-19 C

The Attempt at a Solution

EA = Q(V_gs/V_s) = E(4pi*r^2) = ρ(4/3)pi*r^3
so E = (1/3)ρ

F = qE
F = ((1.6*10^-19)(2*10^-3))/3

This is incorrect though and I'm not sure what the solution is. I'm new to Gauss' law and most likely not understanding these equations so any tips/links would be great.

 

[collinsmark]

Homework Statement

The charge density of a spherically symmetric and uniform distribution of charge is ρ = 2*10^-3 C/m^3. An electron is released one centimeter from the center. Find the magnitude and direction of the force onto the electron.

Homework Equations

charge density = Q/V = Q/[(4/3)pi*r^3]

EA = Q(V_gs/V_s)
where E is the electric field, A is the area of sphere, Q is the charge of field, and the ratio is the volume of gaussian surface divided by the volume of entire surface.

Elementary charge constant for an electron = -1.602 * 10^-19 C

The Attempt at a Solution

EA = Q(V_gs/V_s) = E(4pi*r^2) = ρ(4/3)pi*r^3
so E = (1/3)ρ

F = qE
F = ((1.6*10^-19)(2*10^-3))/3

This is incorrect though and I'm not sure what the solution is. I'm new to Gauss' law and most likely not understanding these equations so any tips/links would be great.

I think there may be a couple of things here.

Gauss' law is

\oint_{Surface} \vec E \cdot \vec{dA}= \frac{Q_{enc}}{\varepsilon_0}

[Edit: where Q_{enc} is the charge enclosed within the Gaussian surface.]

if you are using SI units. If you are not using SI units, then never-mind about this.

Anyway, on the left hand side of the equation, since there is spherical distribution, the magnitude of the electric field at the Gaussian surface is constant, you you can evaluate the integral as simply EA. That you have done; so far so good.

But on the right hand side of the equation, you left out an \varepsilon_0 somewhere. That's of course, if you are using SI units.

Secondly, in your
EA = Q(V_gs/V_s) = E(4pi*r^2) = ρ(4/3)pi*r^3
equation, there is an r² on the left hand side, and an r³ on the right hand side. So the rs don't completely cancel. So there should be an r in there somewhere too.

 

[ichivictus]

Ah I see. Here's my new work.

EA = E4pir^2 = Q/ε

so E = Qr/(4*pi*ε*R^3)
Q/(4piR^3)= charge density

E = ρr/ε

And if F = E*q

F = ρrq/ε = (2*10^-3)(.01)(-1.602 * 10^-19)/(8.85*10^-12) = -3.62*10^-13 Nt

This is also wrong, answer should be in a lower magnitude around 10^-17. I've tried many ways and keep getting so many different results all over the place.

 

[collinsmark]

You left out a '3' in the denominator this time.

Figure out how that got left out and then try plugging the numbers back in again. Then it should work out.

 

[ichivictus]

Ah duh.

So EA = Q/ε = ρv/ε
A cancels out v except leaves r/3 on the right.
E = ρr/3ε
Eq = F = ρqr/3ε

Plugging the numbers in gets me closer, but not quite there. Perhaps there's a typo in the homework thing. But now I finally understand and I found a similar problem in my book and it matches up. Thank you for your help!