Confused about Superposition of Electric Field

[FCBarca1990]

Hey guys,

I'm a bit confused by the superposition of electric fields; in the Princeton Review Physical Sciences test prep book, it claims that E(net) = E(+) + E(-)

However, this doesn't make a whole lot of sense to me since, if you have two sources of equal and opposite charge, the net electric field halfway between the charges would be:

E(net) = KQ/r^2 + K(-Q)/r^2 = 0

*with r = total distance / 2

However, from a basic understanding of the electric field, we know that it would actually be two times the electric field generated by either source, as shown by the electric field lines:

(+) ----------------------> (-)

Can anyone explain this? It shouldn't be a huge deal since I'm assuming I can use intuition and a more conceptual understanding of the electric field to solve any problem on the MCAT, but it would be good to know how to sensibly calculate it mathematically.

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[FCBarca1990]

Bump....

 

[loua0731]

Remember that electric fields are vectors, so you can't just add the magnitudes like you would for electric potential energy. So determine the values for the field generated by each charge (and take into account the direction, since they're vectors), and add them like you would any other vectors, i.e. Pythagorean Theorem.

 

[FCBarca1990]

Yes, but in my example above, both points lie on the x-axis and have no Y coordinate. My question is how to distinguish which way the vector arrow points based on the numerical value provided by the Electric Field equation.

It seems as though for E = kQ/r^2, a positive charge and thus a positive value substituted for Q points in the same direction of as a negative charge and negative value substituted into Q.

So, if an electric field value is negative, how can we determine in which direction it points without knowing whether the source is positively or negatively charge?

 

[loua0731]

Yes, but in my example above, both points lie on the x-axis and have no Y coordinate. My question is how to distinguish which way the vector arrow points based on the numerical value provided by the Electric Field equation.

It seems as though for E = kQ/r^2, a positive charge and thus a positive value substituted for Q points in the same direction of as a negative charge and negative value substituted into Q.

So, if an electric field value is negative, how can we determine in which direction it points without knowing whether the source is positively or negatively charge?

If the overall electric field value is negative, then the field will point in the direction controlled by the negative source charge.

Let's say you have two opposite charges of equal sign, separated by a distance r:

(+Q)-------r-------(-Q)

And the question asks you to find the net electric field at a point halfway between both charges.

The magnitude of each field generated by the charges will be equal:
positive charge: kQ/(r/2)^2
negative charge: kQ/(r/2)^2

However, the field will not cancel out to zero, because the field vectors will point in the following direction:

(+Q)------->r------->(-Q)

So the net electric field will be 2kq/(r/2)^2, pointing toward the negative charge.

I hope that helps and answers your question!

 

[FCBarca1990]

Hmm...I guess I should clarify; is there any way to determine the direction of the electric field without knowing the identity of the source charges?

If you are given two different values for the magnitude of two electric field vectors, how do you determine whether they add (as you mentioned above) or subtract without seeing a picture of the sources and their respective charges?

 

[LostInStudy]

I've never seen that situation happen. There is no way to know. They usually give you direction or the identity of the charge.

-LIS

 

[loua0731]

Hmm...I guess I should clarify; is there any way to determine the direction of the electric field without knowing the identity of the source charges?

If you are given two different values for the magnitude of two electric field vectors, how do you determine whether they add (as you mentioned above) or subtract without seeing a picture of the sources and their respective charges?

I think the only time you can tell the net direction of electric field is if you have a positive and negative source charge--because no matter what, the net field will point toward the negative charge. Other than that, I don't think you can determine the net direction unless they tell you what the magnitudes of the source charges are. But if they expect you to come up with a vector direction, I would just go ahead and assume that they're both positive source charges. So assuming that they're both positive, and have a magnitude difference of 2, for instance:

(+Q)-----r-----(+2Q) ... and let's say the point of interest is halfway between the two charges..

the magnitudes of each would be:
+Q= kQ/(r/2)^2
+2Q= 2kQ/(r/2)^2

Now, because both source charges are positive...the field vectors would look something like:

(+Q)----><--------(+2Q) [I drew the line pointing to the left longer because of the larger electric field generated by the +2Q charge]

So you need to subtract in this case...and the net field would point to the left due to the second source charge's larger magnitude.

So basically, do not assign +/- signs to the fields for each individual charge based on the charge's sign--I think this might be what's confusing you.
Draw a picture first, and draw arrows pointing in the designated directions, then from that you can determine if you need to add or subtract. If they're both of one sign, you will have to subtract (if the point of interest is in between both charges). If they're opposite in sign, then you will add...and the net direction will be toward the negative charge.

Any question SHOULD at least give you relative charge magnitudes and signs, though.

I hope that clarified it for you!

 

[FCBarca1990]

Okay...I got it. I'll just assume that a picture or some kind of explanation of the source charges will always be associated with the problem. Silly sign conventions....

Thanks for the help! 🙂

 

[LostInStudy]

Okay...I got it. I'll just assume that a picture or some kind of explanation of the source charges will always be associated with the problem. Silly sign conventions....

Thanks for the help! 🙂

Exactly.

-LIS