Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Draw a negatively charged plate A and a positively charged plate B with an equal charge. Draw the field. (the image is the given answer)

I however do not understand why this is. Is there any reason why the arrows between the plates are not placed on the same 'height' but on 2 different heights? Also, the field lines above B and below A are going into 2 directions (which is linked with my previous question), which to me seems weird. What my idea of a correct answer would be: Between the plates all the fields have arrows (on the same height) pointing to A, above plate B all the arrows point upward and below field A all arrows point downwards (however, not sure about the last statement since the field not between the plates is 0 I'd say). I have just learned about fields and field lines, and I still find them hard to grasp, so pardon me for this elementary question. Also, is it a coincidence that the field lines seem to go through the $+$'es perfectly but are not aligned perfectly with the $-$'es?

enter image description here

side note: Before this question you are asked to only draw a field A, it is only after you've answered that they ask you to draw B in the same picture too. This might have to do with the way it is drawn in this picture.

share|improve this question
    
The diagram is confusing. It is drawing two sets of field lines: one set due to plate A (as if plate B didn't exist) and another due to plate B (as if plate A didn't exist). It is not showing the total field. This doesn't represent the total field if both plates are present! –  Michael Brown Jan 8 '13 at 15:10
    
@MichaelBrown How would the total field deviate from this one? Is my guess correct? Or wouldn't you have to draw field lines above B and below A (since those compensate eachother)? –  user14445 Jan 8 '13 at 15:11
    
The electric field is a vector field $\vec{E}$: it has a magnitude and direction. If a charge distribution $A$ produces a field $\vec{E}_A$ and charge $B$ produces $\vec{E}_B$ the total field is the vector sum $\vec{E}=\vec{E}_A + \vec{E}_B$. In this particular example the fields reinforce between the plates (same direction) and cancel outside of the plates (opposite direction). –  Michael Brown Jan 8 '13 at 15:17
    
@MichaelBrown You should gather those comments into an answer, since there is not much more to say. Otherwise I'd answer and maybe surpass you in rep earned this year to date ;) –  Chris White Jan 8 '13 at 15:30
    
@ChrisWhite Bah, you're already well ahead of me in total and you're sure to widen the gap when my holidays are over. :) –  Michael Brown Jan 8 '13 at 15:49
add comment

1 Answer

up vote 4 down vote accepted

(as per Chris White's suggestion)

The diagram is confusing. It is drawing two sets of field lines: one set due to plate A (as if plate B didn't exist) and another due to plate B (as if plate A didn't exist). It is not showing the total field. This doesn't represent the total field if both plates are present!

The electric field is a vector field $\vec{E}$: it has a magnitude and direction. If a charge distribution A produces a field $\vec{E}_A$ and charge B produces $\vec{E}_B$ the total field is the vector sum $\vec{E}=\vec{E}_A+\vec{E}_B$. In this particular example the fields reinforce between the plates (same direction) and cancel outside of the plates (opposite direction).

share|improve this answer
    
Thank you for this answer –  user14445 Jan 8 '13 at 15:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.