**Problem Statement:**

The plates of a parallel plate capacitor have an area of 400 cm^{2} and they are separated by a distance d = 4 mm. The capacitor is charged with a battery of voltage ΔV = 220 V and later disconnected from the battery. Calculate the electric field, the surface charge density σ, the capacitance C, the charge q and the energy U stored in the capacitor.

__Givens__: ε_{0} = 8.854 10^{-12} C^{2} / N m^{2}

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**Solution:**

In this problem we are going to use the expression of the electric field due to a parallel plate capacitor like the one represented in the next figure.

The electric potential difference between its plates is given by:

And from the previous expression we can deduce the surface charge density of the plates because we know the distance between them as well as their area:

After substituting the numerical values we get:

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The surface charge density is equal to the charge per unit surface area:

Therefore, the charge of each plate (one positive and the other one negative) expressed in SI units is:

The magnitude of the electric field inside the capacitor (represented by the field lines in the upper figure) is given by:

The capacitance of the capacitor is the ratio between the charge and the electric potential difference between the plates:

One farad (the SI unit of capacitance) is a very large value of capacitance; therefore, the capacitance of a capacitor is usually expressed using sub multiples of the farad. You can see the names of the multiples and sub multiples of the SI units in the units of measurement page.

Finally, the energy stored in the capacitor is given by:

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