Easy EM problem

1. Homework Statement
For a time independant or static situation

$$-\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a)$$

a) Show that for any closed path
$$\oint \vec{E} \bullet d\vec{l} = 0$$

b)Using Stokes Theorem show that
$$\oint \vec{E} \bullet d\vec{l} = 0$$ implies $$\vec{\nabla} \times \vec{E} = 0$$

c) $-\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a)$ was derived by integrating $\vec{E} = -\vec{\nabla} \Phi$. Using Cartesian coordiants show that this implies $\vec{\nabla} \times \vec{E} = 0$

2. The attempt at a solution

a) For any closed path the endpoints a and b are the same. So
$$-\int_{a}^{a} \vec{E} \bullet d\vec{l} = \Phi(a) - \Phi(a) = 0$$

Is that it? Just one line?? Is there more i should be considering?

b) Stokes theorem is

$$\int_{S} \left(\nabla \times v) \bullet da = \oint_{P} v \bullet dl$$

so here we already know that
$$\oint \vec{E} \bullet d\vec{l} = 0$$
thus the right hand side is zero

$$\int_{S} \left(\nabla \times E) \bullet da = 0$$
ok since the integral is zero either one of two things - either the integral was over some closed surface S and/or the integrand was zero
or the curl of E is perpendicular to da

how would i prove thi formally??

c) All i do is simply take the curl on both sides of
$\vec{E} = -\vec{\nabla} \Phi$
since the curl of a gradient is zero we prove that it is zero

since the proof asks to use Cartesian coords i guess i have to do something like this

$$- \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right)$$
and prove that the above expression is zero

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cristo
Staff Emeritus
1. Homework Statement
For a time independant or static situation

$$-\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a)$$

a) Show that for any closed path
$$\oint \vec{E} \bullet d\vec{l} = 0$$

b)Using Stokes Theorem show that
$$\oint \vec{E} \bullet d\vec{l} = 0$$ implies $$\vec{\nabla} \times \vec{E} = 0$$

c) $-\int_{a}^{b} \vec{E} \bullet d\vec{l} = \Phi(b) - \Phi(a)$ was derived by integrating $\vec{E} = -\vec{\nabla} \Phi$. Using Cartesian coordiants show that this implies $\vec{\nabla} \times \vec{E} = 0$

2. The attempt at a solution

a) For any closed path the endpoints a and b are the same. So
$$-\int_{a}^{a} \vec{E} \bullet d\vec{l} = \Phi(a) - \Phi(a) = 0$$

Is that it? Just one line?? Is there more i should be considering?
This is correct
b) Stokes theorem is

$$\int_{S} \left(\nabla \times v) \bullet da = \oint_{P} v \bullet dl$$

so here we already know that
$$\oint \vec{E} \bullet d\vec{l} = 0$$
thus the right hand side is zero

$$\int_{S} \left(\nabla \times E) \bullet da = 0$$
ok since the integral is zero either one of two things - either the integral was over some closed surface S and/or the integrand was zero
or the curl of E is perpendicular to da
Well, $$0=\oint \vec{E} \cdot d\vec{l}= \int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}\Rightarrow\int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}=0$$

Since this is for an arbitrary surface, $$\nabla \times \vec{E}=0$$

c) All i do is simply take the curl on both sides of
$\vec{E} = -\vec{\nabla} \Phi$
since the curl of a gradient is zero we prove that it is zero

since the proof asks to use Cartesian coords i guess i have to do something like this

$$- \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right)$$
and prove that the above expression is zero
Yup, try doing that.

Kurdt
Staff Emeritus
Gold Member
Part a is fine. All you need to know about a closed path is that teh start and end points are the same.

Part b there is only one possibility. The integral over a closed surface of a vector area is zero, but since the vector area is dotted into the curl of the electric field that implies what?

Part c have you written out $$- \nabla \times \left(\frac{\partial \Phi}{\partial x} +\frac{\partial \Phi}{\partial y} +\frac{\partial\Phi}{\partial z} \right)$$
in full? Try it and see what cancels.

EDIT: Beaten again. I'm making a habit of this.

Well, $$0=\oint \vec{E} \cdot d\vec{l}= \int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}\Rightarrow\int_{S} (\nabla \times \vec{E}) \cdot d\vect{A}=0$$

Since this is for an arbitrary surface, $$\nabla \times \vec{E}=0$$

but why is the curl of E zero as a result??
i men the integral could be zero if th curl of E is perpendicular to da
or
the integral was taken over a closed region
or
the curl of E was zero

how does one imply the other??

cristo
Staff Emeritus
The result follows from the fact that there is no restriction on the surface. If curlE, when integrated over any surface is zero, then it must follow that curlE = 0.

The result follows from the fact that there is no restriction on the surface. If curlE, when integrated over any surface is zero, then it must follow that curlE = 0.
what do u mean restriction on the surface??

Kurdt
Staff Emeritus
Gold Member
He means the surface can be any surface you like. It is not a defined shape.

what do u mean restriction on the surface??
Note that:

$$\int_{S} (\nabla \times \vec{E}) \cdot d\vect{A} \neq \oint_{S} (\nabla \times \vec{E}) \cdot d\vect{A}$$

in general. Thus, you could choose a closed surface, but you don't have to. It can be any surface, including one that isn't closed. The only way this can be true for any surface is if the curl of E is 0.

This is strongly related to the fact that E produces a conservative force.

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