Finding force from electric potential energy using gradients.

AI Thread Summary
The discussion revolves around calculating the electric force exerted on a point charge in a spherically symmetric but non-uniform charge distribution. The electric potential energy U is provided, and the force F is derived using the gradient of U. The participants clarify the correct form of U and its dimensions, emphasizing that U should have units of Joules. There is a focus on ensuring the derived force has the correct units, leading to discussions about the relationships between variables and the importance of including all constants in calculations. The conversation highlights the need for careful derivation and unit consistency in physics problems.
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Homework Statement



In a certain region, a charge distribution exists that is spherically symmetric but
non-uniform. When a positive point charge q is located at (r,θ,φ) near this charge
distribution, there is a resulting electric potential energy for the system given by:

U(r,θ,φ) = ρ(naught)a^2q/18ε(naught)(1-3((r/a)^2) + 2((r/a)^3) for r ≤ a

and 0 for r ≥ a

where ρo is a constant having units of C/m^3 (volume charge density) and a is a
constant having units of m. Note that there is no θ or φ dependence here since the
charge distribution is spherically symmetric.
Determine the electric force F
exerted on charge q as a function of its location
(r,θ,φ) for:
a) r ≤ a
b) r > a
Check that the units of your answer make sense. (Show your work.)

Homework Equations



F=-∇U
∇U=dU/dr r

The Attempt at a Solution

'

I derived ∇U=dU/dr r from the spherical coordinate gradients and since there is no dependence on phi and theta we will just be using the r vector.

therefore:

d/drρ(naught)a^2q/18ε(naught)(1-3((r/a)^2) + 2((r/a)^3) r

= 6r(r-a)/a^3 r

This is where I don't know what to do, how would I express my answer?
 
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You've calculated grad(U) ... you have the relationship between F and grad(U) ... you are asked for F... what's the problem?

Maybe this will help: You have told us...

$$U(r,\theta,\varphi) = U(r) = \frac{\rho_{0}a^2q}{18\epsilon_0 (1-3\left (\frac{r}{a}\right )^2)} + 2\left (\frac{r}{a}\right )^3 \; : \; r \leq a$$ $$\vec{F} = -\nabla U(r,\theta,\varphi)=-\frac{d}{dr}U(r)\hat{r}$$ $$\frac{d}{dr}U(r)=\frac{6r(r-a)}{a^3}$$... note: I'd check that derivative: I don't see where ##\rho_0##, ##q## and ##\epsilon_0## went for eg.
 
Last edited:
Thank you for clearing things up.

At this point I'm just having trouble proving that my answer will be in Newtons (correct units). Which means I probably derived wrong.
 
In fact, [a]=L so your answer has dimensions of inverse-length.
You should go over the derivation.

Is the expression for U(r) (above) the same as what you are given?
The second term 2((r/a)^3) is dimensionless ... what dimensions should U have?
 
Sorry I must have mistyped the equation in the first place.

It should have been:

(ρ(naught)a^2q/18ε(naught)) (1-3((r/a)^2) + 2((r/a)^3)

Hopefully that makes it more clear.

The units for U should be Joules I believe (could be mistaken)
 
(ρ(naught)a^2q/18ε(naught)) (1-3((r/a)^2) + 2((r/a)^3)
Here, let me help with that... $$U(r)= \frac{\rho_0 a^2 q}{18\epsilon_0} \left ( 1-3\left( \frac{r}{a} \right )^2 +2\left ( \frac{r}{a} \right )^3 \right )$$... LaTeX is totally worth the effort of learning it.

To proceed, I would either change the variable, say z=r/a so $$\frac{dU}{dr}=\frac{1}{a}\frac{dU}{dz}$$... or just rearrange $$U(r)= \frac{\rho_0 q}{18\epsilon_0 a} \left ( a^3 -3ar^2+2r^3 \right )$$... you should be able to see right away that you don't have enough constants in your answer... probably you just forgot to put them back at the end.

BTW: "dimensions" is different from "units" ... electric potential has units of "Volts" (energy per unit charge) ... charge is dimensionless so U has dimensions of energy - which would be M.L2T-2.
 
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