# Electric Field in a point

• I

## Main Question or Discussion Point

Hello.

I have a doubt regarding the direccion of the electric field in a point from a source charges. Let be a charge in the left of the plane. In a point to the right of the charge, the direction of the electric field vector is to the right. If the charge is negative, the direction of the electric field vector is to the left.

Why is this? It's just by plain convention?

Thanks

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BvU
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Electric field is simply force per charge. So if equal charges repel each other (and they do), the field to the right of a positive charge points to the right.

But, the electric chage in a point in the plane: in the point there is no charge at all! So, how does the principle of equal charges applies here?

BvU
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Same as with the gravitational field (which, however, is exclusively attractive): the field is there even when there is no mass to pull on.
In the point there is no charge at all, ...
but a test charge $\ q_{\rm test} \$ would experience a force $\ \vec F = \vec E\over \displaystyle q_{\rm test} } \$ and that determines the electric field.

Same as with the gravitational field (which, however, is exclusively attractive): the field is there even when there is no mass to pull on.
but a test charge $\ q_{\rm test} \$ would experience a force $\ \vec F = \vec E\over \displaystyle q_{\rm test} } \$ and that determines the electric field.
Only if the test charge is positive. Why it can´t be negative? Just for convention?

BvU
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Only if the test charge is positive. Why it can´t be negative? Just for convention?
No. Also if the test charge is negative.

BvU
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Except I should have written $$\ \vec F = q_{\rm test} \vec E$$ sorry for the mistypo

No. Also if the test charge is negative.
But, with a negative test charge, the electric field vector shoudn't point to the left?

BvU
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It doesn't. The force $\vec F$ points to the left, but the electric field $\vec E = \displaystyle {\vec F \over q_{\rm test}}\$ still points to the right

It doesn't. The force $\vec F$ points to the left, but the electric field $\vec E = \displaystyle {\vec F \over q_{\rm test}}\$ still points to the right
I don´t see any reason of the WHY a positive charge generate an electric field outward the positive charge, except for a convention. If we assume that the electric field of a positive charge points inward and a negative charge points outward, it would be just the same as is assumed in the textbooks.

BvU
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2019 Award
I don´t see any reason of the WHY a positive charge generate an electric field outward the positive charge, except for a convention. If we assume that the electric field of a positive charge points inward and a negative charge points outward, it would be just the same as is assumed in the textbooks.
Granted. We could also agree that an electron has a positive charge. It wouldn't change a thing - physically. But confusion is guaranteed.

Further on in the curriculum, the electric field will be minus the gradient (measure of change) of the potential. And it takes work $qV$ to move a charge $q$ from infinity to a place with potential $V$. Same issue.

• carlitos_30
Dale
Mentor
I don´t see any reason of the WHY a positive charge generate an electric field outward the positive charge, except for a convention. If we assume that the electric field of a positive charge points inward and a negative charge points outward, it would be just the same as is assumed in the textbooks.
Yes, it is just a convention. We could instead use the convention that protons are negative.

• carlitos_30
Thanks both for your time and help.