Pentafem said:
Ok, thanks, I think I got this now.
Just to check that I got the concept. If we take x1=-1.0m and x2=2.0m, we would get
First exercise: ( V(x) = 6000/x )
v1=-6000V
v2=-3000V, where V2 has the higher potential
Be careful here. Let's go back to that original integral.
\Delta V = -\int \vec E \cdot \vec{dl}
Above, \vec E and \vec {dl} are both vectors with magnitude and direction. But the dot product, \vec E \cdot \vec{dl}, is a scalar!
What does this mean? Well let's apply it to our particular problem here.
V = -\int_r ^{\infty} \frac{b}{r^2} \hat r \cdot \vec{dr}
But note that for our particular problem with a point charge at the origin, \vec{dr} always points in the same direction as \hat r. Rewriting the equation,
V = -\int_r ^{\infty} \frac{b}{r^2} \hat r \cdot dr \ \hat r = -\int_r ^{\infty} \frac{b}{r^2} dr \ \hat r \cdot \hat r
Noting that \hat r \cdot \hat r = 1,
V = -\int_r ^{\infty} \frac{b}{r^2} dr
And we're left only with scalars. Now r is the
distance from the charge at the origin to the positive test charge: not the displacement. In other words, r is now positive scalar.
Evaluating the integral,
V = -\int_r ^{\infty} \frac{b}{r^2} dr = \frac{b}{r}
So when you plug in a value for
x to find the potential, you need to plug in the scalar distance, not the vector displacement.
(In other words, the potential at x = -1, is V(-1) = 6000/1. A positive result [since
b is positive (in this case) and
r is a positive scalar].)
The fact that the
r should be treated as a positive distance, such that you can ignore the sign of
x, is a result of this specific problem. If E was defined some other way, you can't automatically assume that
x is equal to some scalar distance. It's just that you can for this
specific problem.
(And by the way, I just want to reiterate, when the coursework stated "Because V1=V2 + 3.00 kV , the point at x = 2.00 m is at the higher potential," I'm pretty sure that is a mistake. It's obvious that
V1 is 3.00 kV larger than
V2, so the point at
x = 1.00 m has the higher potential.)
Second exercise: ( -1/4 2000*x^4 )
V1=0.5kV
V2=8kV, where V2 has the higher potential, right?
For the other exercise we have
\vec E = b x^3 \hat \imath
Notice that
x here is not defined as a distance such as
r was in the last problem. So later on, when when we plug in numbers for negative values of
x, we need to keep the negative. But we'll come to that in a second.
So we have the integral,
\Delta V = -\int bx^3 \hat \imath \cdot dx \hat \imath = -\int bx^3 dx
But we haven't picked the limits of integration yet. It doesn't make sense to integrate from
x to ∞, similar to what we did for the last problem. The electric field blows up as
x approaches ∞, so those are not good limits in this case.
What you have done is to find the potential at point
x, relative to the origin. Which is a fine thing to do, I suppose. (Or we could calculate the path from
x1 to
x2 as your coursework did. And your coursework's choice of limits is probably a better way, and more direct. But either way works.)
\Delta V = = -\int_0^x bx^3 dx = -\frac{1}{4}bx^4
Now remember, in this particular problem we need to keep the sign for
x when evaluating the potential. That's because
x ≠ |
r| like it was in the last problem. But it just so happens in
this problem that
x4 is always positive. That means the potential, relative to the origin, is always negative.
So getting back your evaluation,
Second exercise: ( -1/4 2000*x^4 )
V1=0.5kV
V2=8kV, where V2 has the higher potential, right?
You forgot the negative sign.
V1 = -0.5 kV
V2 = -8 kV
And
V1 has the higher potential.
Once again it looks like a mistake in the coursework. Take a look at the line right above the final answer in your coursework pdf. It states,
"Simplify to obtain: V_2 - V_1 = -7.5 \ \mathrm{kV}"
which is correct. But then on the very next line it states
"Because V_2 = V_1 + 7.5 \ \mathrm{kV} ..."
which is simply wrong. Any basic algebra from the previous equation shows that
V1 has the higher potential.
I appreciate your frustration with this. It looks to me as though your coursework pdf has a few mistakes in it.
(I wouldn't use the term 'random' charge though. How about 'arbitrary', positive test charge.)
