A parallel-plate capacitor has a capacitance of 100 pF, a plate area of 80 cm^2, and a mica die

A parallel-plate capacitor has a capacitance of 100 pF, a...

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by an insulating material or vacuum. It is one of the simplest capacitor designs and its capacitance is determined by the area of the plates, the distance between them, and the medium’s properties. This concept is foundational in understanding how electrostatic energy is stored in electric fields between the plates.

Dielectric Constant

The dielectric constant, or relative permittivity, is a measure of a material's ability to reduce the effective electric field within it compared to a vacuum. It influences the capacitance of a capacitor by effectively increasing it when a dielectric material is inserted between the plates, reflecting how the material polarizes in response to the field.

Capacitance with a Dielectric

Capacitance is the measure of a capacitor's ability to store charge per unit voltage and is affected by the introduction of a dielectric. In the presence of a dielectric, the capacitance is increased by a factor equal to the dielectric constant. This concept connects the physical attributes of the capacitor’s geometry with the properties of the insulating medium.

Electric Field in a Dielectric

The electric field in a dielectric medium is a key concept in electromagnetism, as it determines the force experienced by charges within the material. In a capacitor filled with a dielectric, the relationship between the applied voltage and the plate separation still defines the average electric field, but the dielectric influences aspects like energy density and field distribution through polarization.

Free Charge on Capacitor Plates

Free charge refers to the charge that is physically supplied to the capacitor plates from an external source, which is responsible for creating the electric field within the capacitor. This concept is essential in linking the physical configuration of the capacitor to measurable electrical quantities like stored energy and potential difference.

Induced Surface Charge (Polarization)

When a dielectric is placed in an external electric field, its molecules become polarized, resulting in induced surface charges on the boundaries of the dielectric. These bound charges are not free to move like conduction electrons but create an internal field that partially cancels the applied field. Understanding induced surface charge is crucial for analyzing how dielectrics modify the effective electric field and the overall behavior of capacitors.

Transcript

00:01 In this question we have been given a parallel plate capacitor and it has been given that the capacitance of this parallel plate capacitor is 100 picofarets and the area of the plates of the capacitor is given as 80 into 10 raised to the power minus 4 meters squared.

00:19 The distance between the capacitors is not been given but we have the dielectric constant which is placed inside the mica.

00:27 So, the constant is given to us as 5 .4.

00:30 And the voltage across which the plates are connected is 85 volts.

00:36 So the first thing that we have to calculate in this question here is the electric field, the magnitude of electric field, e, inside the mica.

00:46 So here this can be easily written as the formula, direct formula, q upon a epsilon knot.

00:54 And we'll put an extra k over here because there's a dialect.

00:57 Constant now putting in the values we know q is equal to cv and this remains a epsilon knot k so on putting in the values so c is 100 picofadts so it's 100 into 10 raised to the power minus 12 and v is given to us as 85 and the value of a which is the plates area is given as 80 into 10 raised to the power minus 4 and epsilon not is the standard value 8 .85 into 10 raised to the power minus 12 and the value of k here, the dialectic constant of mica, is given to us as 5 .4.

01:34 So on calculating, this value of e comes out to be 22 ,000 volts per meter.

01:43 This is the unit for electric field.

01:46 Now let's move on to the second part of the question.

01:49 In the second part, they have asked us to calculate the magnitude of free charge on the plates.

01:54 So this can directly be calculated the formula q is equal to cv.

02:00 So we know c here is given to us as 100 picofarets...

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