# I Field between Parallel Plates in a Capacitor

Tags:
1. May 28, 2017

### Buffu

I guess it is a trivial fact that field must be $(\phi_1 - \phi_2)/s$ but I don't get how ? is there a derivation for it ?[/QUOTE]

2. May 29, 2017

### Staff: Mentor

The difference in electric potential (voltage) between the plates is equal to the work done on a test charge to move it from one plate to the other divided by the charge.
The equation for this is: $V=\frac{Fd}{q}$, where $V=ΔΦ=Φ_1-Φ_2$

A static electric field is defined as: $E=\frac{F}{q}$

Substituting $E$ into the first equation: $V=Ed$

Rearranging: $E=\frac{V}{d}$, or $E=\frac{V}{s}=\frac{Φ_1-Φ_2}{s}$

If the voltage between the plates is constant, then the magnitude of the electric field is inversely proportional to the separation between the plates.

3. May 30, 2017

### Buffu

Why is $F$ constant ?

4. May 30, 2017

### Chandra Prayaga

If the voltage between the plates is held constant (this is in your hands), then the electric field between the plates is constant (and uniform), and therefore the force on a point charge q (test charge) between the plates is also constant.

5. May 31, 2017

### vanhees71

It's a solution of Maxwell's equations for infinitely extended plates, i.e., for the field not too close to the boundaries and for the distance between the plates small compared to the extension of the plates.

Now take the case of two infinite plates parallel to the $xy$ plane of a Caratesian coordinate system, one at $z=0$ and one at $z=d$. Obviously the solution is symmetric under translations in $x$ and $y$ direction. Thus the potential should be a function of $z$ only. There are no charges anywhere, and thus
$$\Delta \phi=\phi'=0.$$
So you have
$$\phi(z)=A+B z$$
with $A$ and $B$ constants. Obviously these constants can take different values inside and outside the plates. Since the potential should stay finite in this configuration, you have $B=0$ for $z<0$ and for $z>d$. The overall constant is arbitrary, and we can choose it to be such that $\phi(0)=0$. Then you have
$$\phi(z)=B z \quad \text{for} \quad 0 < z < d.$$
At $z=d$ you have $\phi(d)=U$, where $U$ is the given voltage difference between the plates, which leads to $B=U/d$ and thus
$$\phi(z)=U \frac{z}{d} \quad \text{for} \quad 0 < z < d.$$
The electric field is
$$\vec{E}=-\vec{\nabla} \phi=-\frac{U}{d} \vec{e}_z \quad \text{for} \quad 0<z<d, \quad \vec{E}=0 \quad \text{everywhere else}.$$
At the upper plate the normal component $E_z$ makes a jump of size $\sigma=U/d$, and $\sigma$ is the surface charge. At the lower plate you get $\sigma'=-U/d$.

For a finite but large plate you have $\sigma=Q/A$ and thus $U/d=Q/A$ or $Q=A U/d$, i.e., the capacitance is $C=A/d$. If there's a dielectric inside, you have $C=\epsilon A/d$, where $\epsilon$ is the zero-frequency permittivity of the dielectric.

6. May 31, 2017

### Buffu

Thank you for the derivation, I think this was a really nice derivation :).

Sorry but I fail to grasp this. If we did not take $B = 0$ for $z \in \Bbb R - [0, d]$ then why does potential is infinite ?

7. Jun 1, 2017

### vanhees71

because then it gets infinite for $z \rightarrow \infty$.

8. Jun 1, 2017

### Buffu

Does it matter ? I mean we are only concerned with potential between the plates whatever be its value outside the plates ?

9. Jun 1, 2017

### vanhees71

The point is to fulfill the boundary conditions at the plate to fully determine the potential. The boundary condition and infinity makes the electric field vanish outside of the plates, and this finally leads to the determination of the surface-charge density which permits to get the capacitance.