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it is said that the electric potential at the center of uniform electric field is zero. my question is that why is it zero? electric potential is the work done per unit charge.

$V = W/q$

and this work is continuously done on a positive test charge if it (charge) is placed in the electric field.

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I can't understand it quite correctly. Could you re your sentence : "The electric potential at the center of uniform electric field is zero"..? I don't know whether an uniform electric field has a center –  Waffle's Crazy Peanut Nov 3 '12 at 16:06
    
by center, i mean mid point of the distance between the positive and negative plate. –  Rafique Nov 3 '12 at 16:09
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The only potential differences matter in Electromagnetism.

So in the case of a uniform electric field along the x direction produced by 2 plates. With the negative one being left and positive one being right. Now I will assume your plates are infinitely long in the Y,Z plane.

In the region in between the plates. $\phi = - k x +c$.

$E = -\nabla \phi = k $

Here $\phi$ is the electric potential. K is the constant electric field, and c is any arbitrary constant.

Now coming to your question. You say that electric potential at the center of a uniform potential is constant. Now in principle one can shift the potential by any constant. and it would still be alright. It is conventionally taken to be zero at infinity.

If you impose this convention, The potential outside the plates is zero. There is sharp jump in the Potential at the Plates to account for sudden change in charge. This jump is negetive on the positive plate and positive for the Negetive plate. Since the charges are equal The jumps must be equal in magnitude. The electric potential will begin to fall as you move towards the positive plate linearly. The symmetry of the situation allows us to establish that he mid-point will have zero potential.

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Consider two equal and opposite charges ($+q$ & $-q$) in space separated by a distance $2r$. An uniform electric field would exist between both acting from $+q$ to $-q$. The first thing is, Electric potential is a scalar quantity whereas Electric field is a vector..! In other words, Electric field is a measure of how the electric potential changes quickly with distance (gradient or the first derivative).

The electric potential at a distance $r$ from $+q$ would be $V_1=\frac{kq}r$

Now, the electric potential at a distance $r$ from $-q$ is $V_2=-\frac{kq}{r}$

The net (effective) potential at midpoint ($r$) is $V=V_1+V_2=0$

In case of Electric field, it is non-zero. Because, we would specify the direction only...


Regarding your case, A test (point) charge not necessarily positive. It's just to indicate the existence of an electric field. In presence of a charge, the test charge would experience a force. That's all :-)

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electric potential can also be defined as the energy of a charged particle at a certain point (let this certain point be the center). now, why wouldn't a test charge have energy at the center? work is done on the charge by the electric field to bring it from one of the plates to the center. i understood it mathematically but i am confused theoratically. –  Rafique Nov 3 '12 at 16:53
    
@CrazyBuddy - I have a question... the field is defined as the gradient of the potential ($E = -\nabla V$). So if the potential is zero at that point, wouldn't the field also be zero? –  Kitchi Nov 3 '12 at 17:01
    
@Kitchi: Hello Kitchi. There's a misunderstanding. The potential gradient is actually the change of potential with distance. And $E=-dV/dx$ and not simply $E=-dV$. Hence, Electric field is zero within a small area only if the change of potential with distance is zero at that area..! Think it of a bit :-) –  Waffle's Crazy Peanut Nov 3 '12 at 17:09
    
@MuhammadRafique: Hello Muhammad, It seems like you're really having problem with the test charge. If you really have, forget about it..! (That's only given for definition by these geniuses). My thought is that, Test charges are necessary only for forces (not for energies). So, better think only of your charge :-) –  Waffle's Crazy Peanut Nov 3 '12 at 17:14
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@MuhammadRafique: No, Muhammad. V may be zero. But, dV don't necessarily have to be zero. Consider three points. At the first point, the potential is -2 V, at the second one - it's 0 V and at the 3rd - 2 V. The points are at unit separation. Now, The electric potential is zero at second point. But, the difference in potential from 1<->2 or 2<->3 is 2V. Hence, the field is 2 V/m. Did you get it now..? –  Waffle's Crazy Peanut Nov 3 '12 at 17:26
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