A parallel-plate capacitor has capacitance C=12.5 pF when the volume between the plates is fille

A parallel-plate capacitor has capacitance C=12.5 pF when...

A parallel-plate capacitor has capacitance $C=12.5 \mathrm{pF}$ when the volume between the plates is filled with air. The plates are circular, with radius 3.00 $\mathrm{cm}$ . The capacitor is connected to a battery, and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted, the charge on each plate has magnitude 45.0 $\mathrm{pC}$ (a) What is the dielectric constant $K$ of the dielectric? (b) What is the potential difference between the plates before and after the dielectric has been inserted? (c) What is the electric field at a point midway between the plates before and after the dielectric has been inserted?

Key Concepts

Electric Field in a Capacitor

The electric field between the plates of a capacitor is related to the potential difference by the equation E = V/d, where d is the separation between the plates. The presence of a dielectric reduces the effective electric field inside the capacitor because the dielectric becomes polarized, effectively reducing the field within it. This concept is important for determining how the electric properties of a capacitor change when a dielectric material is inserted.

Dielectric Constant

The dielectric constant (often denoted by K) quantifies the ability of a dielectric material to increase the capacitance of a capacitor compared to the capacitance with air or vacuum as the dielectric. It is a dimensionless number greater than 1, indicating how much the internal electric field is reduced (and therefore the capacitance is increased) when the material is introduced. This concept is fundamental when analyzing the impact of inserting a dielectric slab in a capacitor that is connected to a battery.

Voltage-Charge-Capacitance Relationship

The relationship among voltage, charge, and capacitance is expressed by the equation Q = CV. When a capacitor is connected to a battery, the voltage remains constant, so any change in capacitance due to modifications such as inserting a dielectric will result in a proportional change in the stored charge. This principle is central to understanding the behavior of the system when the dielectric is introduced.

Parallel-Plate Capacitor

A parallel?plate capacitor consists of two conducting plates separated by a dielectric (which may be a vacuum, air, or another material). The geometry of the plates (area and separation) directly influences its ability to store charge, and thus its capacitance. In problems of this nature, understanding the basic structure and function of such a capacitor is crucial, particularly how its physical dimensions and the medium between the plates come into play.

Capacitance

Capacitance is the measure of a capacitor’s ability to store electrical charge per unit voltage, defined by the relation C = Q/V. It depends on both the geometry of the capacitor (i.e., the area of the plates and the distance between them) and the properties of the dielectric material between them. In many problems, this concept is key to connecting charge, voltage, and stored energy, and is influenced by modifications such as inserting a dielectric.

Transcript

00:02 Hi, in the given problem, we are discussing a parallel plate capacitor before inserting the dielectric slab between its plates and after inserting the dielectric slab between its plates.

00:16 Its initial capacitance before inserting the dialectic slab has been given as 12 .5 pico -farratt or we can say this is 12 .5 into 10 to power minus 12 ferret.

00:33 This capacitor is made using the circular plates and the radius of these circular plates has been given as 3 .00 centimeter or we can say this is 3 .00 into 10 dash to power minus 2 meter.

00:55 Now the charge over the plates initial charge with the out the dielectric slab between the plates.

01:04 The charge was 25 .0 pico -culum and after inserting the dielectric slab, this charge becomes q -dash is equal to 45 .0 pico -cula.

01:21 The battery remains connected between the plates of capacitor in the act of inserting the dialectics lap between the plates.

01:30 Now, in the first part of the problem, we have to find the dialectic constant of the medium.

01:37 As we know, on inserting the dialectic slab between the plates of the capacitor, when the battery still remains connected, the charge becomes k times.

01:48 So we can say the dialectic constant k is nothing but equals to q -by -q.

01:54 So putting these values here, 45 .0, pico -pulam divided by 25.

02:00 0 p .0 p .coma coulame we get the expression 4 we get the value for this directory constant as 1 .8 and this is the answer for the first part of the problem now in the second part of the problem we have to find the potential applied between the plates of this capacitor for which we will use the simple relation between potential and charge and capacitance which is v equals to q by c...

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