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Can an Electric Field with field lines Like So Exist:

One Of my friends said it couldn't as the field lines here are not conservative ; so it cannot exist ; Is he right?

Or can it be made to exist

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closed as unclear what you're asking by Emilio Pisanty, Brandon Enright, Chris White, tpg2114, John Rennie Feb 3 '14 at 11:18

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

Dont the Electric Field Lines mean the same thing? – The-Ever-Kid May 10 '12 at 12:37
Yep. "Lines" is missing in your question. EDIT: I see, you have it in title, but you dont have it in the text. Never mind. – Pygmalion May 10 '12 at 12:39
"Existence Of Electric Field Lines" & "as the field lines here" Ive used the word like im editing it. – The-Ever-Kid May 10 '12 at 12:40
OK, fine, I somehow missed title reading text :) – Pygmalion May 10 '12 at 12:41
The link to this image is broken. Please replace it with an image hosted at the standard StackExchange image hosting service by using the image upload button on the question editing window. Or does anyone remember what that picture was? – Emilio Pisanty Feb 3 '14 at 0:46
up vote 4 down vote accepted

Yes, your friend is right. Within electrostatics, an electric field $\vec{E}$ should be curl-free $\vec{\nabla} \times\vec{E}= \vec{0}$. The drawn electric field lines looks like the electric field is of the form

$$ E_x=E_x(y), \qquad E_y=0, \qquad E_z=0, $$

cf. the rule that to depict the magnitude $|\vec{E}|$, a selection of field lines is drawn such that the density of field lines (number of field lines per unit perpendicular area) at any location is proportional to $|\vec{E}|$ at that point. Here the $x$-axis is horizontal, the $y$-axis is vertical, and the $z$-axis perpendicular to the plane.

This is only curl-free if $E_x=E_x(y)$ is independent of $y$, which it isn't on the figure.

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Can we Make Something that generates a field like so 'coz I've Heard that induced electric field is non conservative – The-Ever-Kid May 10 '12 at 12:34
He didn't say the electric field was conservative – Physiks lover May 10 '12 at 18:43
@The-Ever-Kid: Perhaps you are referring to Faraday's induction law. With the help of an appropriate time-varying magnetic field, the electric field lines in the figure may be realized. – Qmechanic May 10 '12 at 18:51
@Physiks lover: Well, indirectly. OP tagged it as electrostatics, which traditionally implies a curlfree electric field. – Qmechanic May 10 '12 at 18:56
+1, although technically if there is nontrivial $z$-dependence, this could be a slice of a zero-curl configuration. (Admittedly one is not given any reason to believe that is the case.) – David Z May 10 '12 at 20:52

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