First in a book I found that the electric field strength $\mathrm{E}$ of a uniform electric field from a charged plate is equal to $\frac{q}{2\times \epsilon_0 \times S}$ where $q$ is the charge of the plate, $\epsilon_0$ is the electrical permitivitty and $S$ is the area of the plate.

My first question is how to find the potential of a point in the uniform field (is there a formula using the electric field strength and may be some distance to the plate) or at least the voltage between two points (again formula using the electric fields strength and the distance between the two points).

Now we are talking about a capacitor. In some sources it is said that one plate has negative electric potential (for example $V$) and the other $-V$. But in other sources one plate is said to have a zero potential and the other to have nonzero. Are these both true? In my opinion the negative plate would make negative potential in the positive plate and vice versa (we can look at the uniform field of the capacitor as made of from the two uniform fields created by the negative and the positive plate). I'm really confused.

Also I want to ask whether there is a formula for finding the electric potential of a point in the capacitor because for the voltage between two points it is $E\times d$ where $d$ is the distance between them. Finally, I have found that the electric potential in the midpoint of the capacitor (i.e. if the distance between the plates is $d$, the distances between this point and the plates is $\frac{d}{2}$).

Thanks in advance.


The relationship between electric field and electric potential is just that the electric field is (minus) the gradient of the potential. Thus in the case of a uniform field extending from a uniformly charged plate (let's call it along the z-axis, with the late in the x,y plane) $$ E_z = - \frac{dV}{dz}$$.

This of course means that $$ V(z) = - \int E_z\ dz = E_z z + V_0,$$ where $E_z$ is the constant field and $V_0$ is an arbitrary constant. What this means is that you can set the zeropoint of the voltage to be anything you like, it is only its gradient that matters. The potential difference simply depends on the difference in $z$ coordinates, $\Delta z$. ie. $V_1 = E_z z_1 + V_0$ and $V_2 = E_z z_2 + V_0$, so the potential difference $V_2- V_1 = E_z \Delta z$.

Therefore the potential at some distance $z$ between the plates of an ideal, parallel-plate capacitor with uniform electric field $E$ and the positively charged plate is at $z=0$ is $E z + V_0$. You can choose $V_0$ to be anything you like, but whatever you choose, the potential difference between the plates $(E d + V_0) - V_0$ is exactly the same.

Thus when you ask what is the potential at a point midway between the plates, all you can really say is what is the potential difference between that point and one of the plates, in this case $Ed/2$.

  • $\begingroup$ Thanks for the answer. The only thing I want to ask is about what the potentials of the plates of a capacitor are equal to. $\endgroup$ – Илиян Йорданов Apr 3 '15 at 20:07
  • $\begingroup$ The potential difference is the charge on the plates divided by the capacitance $=Ed$. You can set the zeropoint of the potential to whatever is convenient. $\endgroup$ – ProfRob Apr 3 '15 at 23:43

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