Electric Field in a Capacitor

[kash25]

Hi,
I have a problem with electric fields in capacitors. Using Gauss' Law, we always choose the Gaussian surface that encloses the positive plate in the capacitor. When we go through field calculations, we get an equation for the electric field and we use that to find voltage, capacitance, and everything else. What I don't understand is how enclosing ONLY the positive plate takes into account the fact that the negative plate is present and that it amplifies the field between the two. In other words, how would our calculations or conclusions be any different if the negative plate were not even present?
Thank you very much for the help

 

[alphysicist]

Hi kash25,

I don't think that's correct. For example, for parallel plates we might use Gauss's law to find the electric field of just the postive plate (giving a result of $E = \sigma/(2\epsilon_0)$ everywhere), but then we would double that to find the electric field of the capacitor between the plates ($E=\sigma/\epsilon_0$) and set the field to be zero outside the plates.

(Between the plates, the fields are in the same direction and so add; outside the plates they are in opposite directions and therefore cancel; also I'm making the usual approximation that the nonuniform field at the ends of the plate can be neglected, etc.)

 

[Moetasim]

!

I understand your confusion and I am happy to provide some clarification on the concept of electric fields in capacitors. First, it is important to understand that a capacitor is made up of two conductive plates separated by a dielectric material. When a voltage is applied, one plate becomes positively charged and the other negatively charged. This creates an electric field between the plates.

Now, when we use Gauss' Law to calculate the electric field, we are considering the overall effect of the charges on both plates. By choosing a Gaussian surface that only encloses the positive plate, we are essentially considering the electric field caused by the positive charge alone. However, this does not mean that the negative plate is not taken into account.

In fact, the presence of the negative plate is crucial in determining the capacitance of the capacitor. The electric field between the plates is amplified due to the presence of the negative charge, leading to a higher capacitance value. If the negative plate were not present, the electric field would be weaker and the capacitance would be lower.

In conclusion, by choosing a Gaussian surface that encloses only the positive plate, we are able to simplify the calculations and still take into account the presence of the negative plate. This approach is widely used and has been validated through experimental observations. I hope this helps clarify your understanding of electric fields in capacitors.