Electric field in a parallel plate capacitor

A capacitor is a device used in electric and electronic circuits to store electrical energy as an electric potential difference (or an electric field). It consists of two electrical conductors (called plates), typically plates, cylinder or sheets, separated by an insulating layer (a void or a dielectric material). A dielectric material is a material that does not allow current to flow and can therefore be used as insulator.

The first capacitor was build in 1745-1746 and consisted of a glass jar covered by metal foil on the inside and outside. It is known as the Leyden jar (or Leiden jar).

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In this page we are going to calculate the electric field in a parallel plate capacitor.

A parallel plate capacitor consists of two metallic plates placed very close to each other and with surface charge densities σ and respectively. The field lines created by the plates are illustrated separately in the next figure.

The magnitude of the electric field due to an infinite thin flat sheet of charge is:

Where ε0 is the vacuum permittivity or electric constant.

The charge density of each plate (with a surface area S) is given by:

The electric field obeys the superposition principle; its value at any point of space is the sum of the electric fields in this point. Therefore, the field on the outside of the two plates is zero and it is twice the field produced individually by each plate between them.

Therefore the magnitude of the electric field inside the capacitor is:


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The capacitance C of a capacitor is defined as the ratio between the absolute value of the plates charge and the electric potential difference between them:

The SI unit of capacitance is the farad (F).

Let’s assume the distance between the capacitor plates to be d as seen in the next figure:

The electric potential difference between them is given by:

If we use the unit vector i to write the electric field vector between the plates, we have:

And we can express the dl vector as:

After substituting both vectors in the integral we obtain:

Finally, the capacitance of the parallel plate capacitor is:


During the charge of a capacitor, a positive charge dq is transferred from the negative plate to the positive one. But, in order to do that, it is necessary to provide a certain amount of energy in the form of work, because if it were not the case, the positive charge would be repelled by the negative plate.

The work done to move the charge dq from the negative to the positive plate is given by:

We integrate between an empty charge and the maximal charge q to obtain:

If we express q as a function of the capacitor’s capacitance we have:

The energy used to charge the capacitor stays stored in it.

Therefore, the energy stored in a charged capacitor is:

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