Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A potential difference is applied between two metal plates that are not parallel. Which diagram shows the electric field between the plates? enter image description here

A is the answer BUT WHY explanation needed !

share|cite|improve this question

closed as too localized by Qmechanic Jun 3 '13 at 19:07

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Each plate is an equipotential surface, so the field lines must be perpendicular to the plates. Also, the field intensity increase where the distance between the plates decreases.(separation between lines indicates field intensity) – Mostafa Jun 3 '13 at 17:29
What level should the answer be? I guess one could do a lot of fancy theoretical electrodynamics with Laplace's equation of $\varphi$ if you need university level. – Martin Ueding Jun 3 '13 at 17:54
nope just such that a AS level could comprehend :D – phantom.omaga Jun 3 '13 at 19:20
up vote 3 down vote accepted

In electrostatics electric fields must be perpendicular to the surface of conductors. Otherwise there would be a component tangential to the surface, which would cause charges to move. The charges would move until they found an equilibrium charge distribution, where there are no more tangential electric fields forcing them to move, i.e. electrostatics.

On the other hand density of field lines describes the strength of the field. And, as you probably know, the electric potential is given by a line integral of the electric fields between two points $V=-\int\mathbf{E}\cdot d\mathbf{l}$. So in order for this integral to give the same answer (the applied voltage) along the upper (longer) and lower (shorter) path the electric field must be stronger at the bottom, hence the increased density of lines.

share|cite|improve this answer

Reference: section 2.2, Griffiths - "Introduction to Electrodynamics"

Field lines have some rules that are often implied and not stated. The arrows describe the direction of the field vector at the point of interest. The magnitude of the field is described by the density of the field lines. Electric field lines begin at positive charges and end at negative charges. Also, electric field lines are perpendicular to the equipotential surfaces of the electric potential. This is because: $$\vec{E}=-\vec{\nabla}\Phi$$

And a property of the gradient field is that it is perpendicular to the equipotential surfaces. This can be seen by observing that the directional derivative of the function $\Phi\left(x,y,z\right)$ , in the direction $\vec{T}$ is: $$\frac{\partial\Phi}{\partial\vec{T}}=\vec{\nabla}\Phi\cdot\vec{T}$$ If we are on an equipotential surface, $\Phi\left(x,y,z\right)=constant$ , then the directional derivative must be zero because there is no change in the function, but then:$$\vec{\nabla}\Phi\cdot\vec{T}=0$$ And the gradient is perpendicular to the equipotential surfaces. In this case the equipotential surfaces are the capacitor plates.

In summary, the field lines must emanate from the positive charges and terminate at the negative charges. Their density must be greater where the plates are closer, because the electric field strength there is greater. The lines must be perpendicular to the capacitor plates (and all the equipotential surfaces you care to draw).

share|cite|improve this answer

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