The plates of a parallel plate capacitor have an area of 90 cm^2 each and are separated by 2.5

The plates of a parallel plate capacitor have an area of 90...

The plates of a parallel plate capacitor have an area of $90 \mathrm{~cm}^{2}$ each and are separated by $2.5 \mathrm{~mm}$. The capacitor is charged by connecting it to a $400 \mathrm{~V}$ supply. (a) How much electrostatic energy is stored by the capacitor? (b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

The plates of a parallel plate capacitor have an area of $90 \mathrm{~cm}^{2}$ each and are separated by $2.5 \mathrm{~mm}$. The capacitor is charged by connecting it to a $400 \mathrm{~V}$ supply.
(a) How much electrostatic energy is stored by the capacitor?
(b) View this energy as stored in the electrostatic field between the plates, and obtain the energy per unit volume $u$. Hence arrive at a relation between $u$ and the magnitude of electric field $E$ between the plates.

Key Concepts

Parallel Plate Capacitor

A parallel plate capacitor consists of two conductive plates separated by a small gap, usually filled with an insulating material (dielectric). Its design leads to a uniform electric field between the plates, and its capacitance is determined by the area of the plates, the separation distance between them, and the permittivity of the medium between them.

Capacitance

Capacitance is a measure of a capacitor's ability to store electric charge per unit voltage. For a parallel plate capacitor, the capacitance is given by the formula C = (??A)/d, where ?? is the permittivity of free space, A is the area of the plates, and d is the separation between them. This concept is crucial for quantifying how much electric potential energy can be stored for a given voltage.

Electrostatic Energy in Capacitors

The electrostatic energy stored in a capacitor represents the work done in charging it, which is stored in the electric field between the plates. It can be expressed as U = ½ CV², where C is the capacitance and V is the voltage applied across the capacitor. This energy quantification is essential in analyzing how capacitors store and release energy.

Electric Field

The electric field in a capacitor is the force per unit charge created in the region between the two plates. In a parallel plate capacitor, this field is approximately uniform and is directly related to the applied voltage and inversely related to the separation distance, expressed as E = V/d. Understanding the electric field is key to linking voltage differences and stored energy.

Energy Density in Electric Fields

Energy density refers to the amount of energy stored per unit volume in an electric field. In free space, the energy density u is given by u = ½ ?? E², indicating that the energy stored in a given volume is directly related to the square of the electric field strength. This concept bridges the energy stored in a capacitor with the spatial distribution of the electric field.

Transcript

00:01 This question belongs to the capacitors in which we have a plate of parallel plate capacitor having area a equals to 90 centimeter square that is equals to 90 multiplied and to the power minus 4 meter square.

00:13 And each plate is separated by a distance d equals to 2 .5 mm that is equals to 2 .5 multiplied by 10 to the power minus 3 meter.

00:23 The capacitor is charged to connecting by it with v equals to 400 volt supply.

00:29 So for the part a, we have to calculate the energy stored in the capacitor that is u.

00:35 So we know that the energy stored in the capacitor u, it is equals to 1 by 2 c v.

00:41 Square where c is the capacitance of the capacitor and v is the potential difference across the capacitor.

00:47 And capacitance of the parallel plate capacitor, it is equals to epsilon naught a by d.

00:52 So substituting values, so epsilon not is 8 .854 multiplied by 10 to the power minus 12, multiplied by area which is 90 multiplied by 10 to the power minus 4 divided by small d which is 2 .5 multiplied by 10 to the power minus 3 from here.

01:09 So capacitance c it is equals to capacity it is equals to the this value.

01:18 So we can use this relation to substitute values here.

01:22 So we will get that u it is equals to 1 by 2 and c which is this complete value.

01:27 So 8 .854 multiplied by 10 to the power minus 12 multiplied by 90 multiplied by 10 to the power minus 4 divided by d which is 2 .5 multiplied by 10 to the power minus 3 multiplied by v which is 400 volts so 400 volt and this whole square.

01:44 So from here after solving we get energy stored in the capacitor u it is equals to 2 .55 multiplied by 10 to the power minus 6 jule.

01:54 So this become the answer for this question okay so this might be able to the power...

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