A parallel-plate capacitor has plates of area 0.12 m^2 and a separation of 1.2 cm . A battery

A parallel-plate capacitor has plates of area 0.12 m^2 and...

A parallel-plate capacitor has plates of area $0.12 \mathrm{~m}^{2}$ and a separation of $1.2 \mathrm{~cm} .$ A battery charges the plates to a potential difference of $120 \mathrm{~V}$ and is then disconnected. A dielectric slab of thickness $4.0 \mathrm{~mm}$ and dielectric constant $4.8$ is then placed symmetrically between the plates. (a) What is the capacitance before the slab is inserted? (b) What is the capacitance with the slab in place? What is the free charge $q$ (c) before and (d) after the slab is inserted? What is the magnitude of the electric field (e) in the space between the plates and dielectric and (f) in the dielectric itself? (g) With the slab in place, what is the potential difference across the plates? (h) How much external work is involved in inserting the slab?

Key Concepts

Energy Storage and Work Considerations

A capacitor stores electrical energy in its electric field, and the stored energy is given by formulas such as U = ½CV² or U = Q²/(2C). Changes in capacitance, as occur when a dielectric is inserted, lead to changes in stored energy. Additionally, the work done by external forces in processes like inserting a dielectric slab is related to these energy differences. This concept is key to understanding the energy balance and the amount of external work required during such processes.

Insertion of Dielectric

Dielectric materials are insulators that, when inserted between the plates of a capacitor, reduce the effective electric field and increase the capacitance by a factor equal to the dielectric constant (?). When a dielectric is partially or completely inserted, the system can often be modeled as a combination of capacitors in series or parallel depending on the configuration. This concept is crucial for understanding the modified capacitance, electric field distribution, and potential difference when a dielectric slab is introduced.

Electric Field in Capacitors

The electric field between the plates of a capacitor is determined by the potential difference and the separation between the plates. In a capacitor partially filled with a dielectric, the presence of regions with different permittivities results in varying electric field strengths, with the dielectric reducing the field within its region. Understanding how to calculate the electric field in each region is essential for analyzing the voltage distribution and overall behavior of the capacitor.

Battery Disconnection and Charge Conservation

When a capacitor is charged by a battery and then disconnected, the charge on the plates remains constant. This is an important consideration because subsequent changes—such as the insertion of a dielectric—alter the capacitance and voltage across the capacitor while the total free charge remains fixed. This conservation of charge underpins the analysis of the capacitor’s behavior after the dielectric is inserted.

Parallel?Plate Capacitor

A parallel?plate capacitor is formed by two conductive plates facing each other, separated by a dielectric (which may be air or another insulating material). The capacitance of such a device depends on the plate area, separation distance, and the permittivity of the medium between the plates. This concept provides the framework for understanding how charge is stored in an electric field between the plates.

Capacitance Calculation

Capacitance is defined as the ratio of the charge stored to the voltage across the capacitor and, for a parallel?plate capacitor without dielectric, is given by C = ??A/d, where A is the area of the plates, d is the separation, and ?? is the vacuum permittivity. This formula is central to determining the initial capacitance before any modifications such as inserting a dielectric are made.

Transcript

00:01 So in this problem, we're considering a capacitor that is charged to a potential difference of 120 volts.

00:10 And then you place a slab of dielectric material between the plates.

00:17 And so we want to start off by trying to figure out what is the capacitance of the capacitor before we add the dielectric.

00:26 Before the dielectric is added, capacitants would be given by epsilon.

00:31 1 .a divided by d.

00:34 And so that's just going to be 8 .85 times 10 to the negative 11 ferrads.

00:43 For part b, we now want to figure out what the capacitance is after you have added in the dielectric.

00:54 And in that case, we want to consider the capacitors as two capacitors in series.

01:01 One that is the air filled capacitor and one that is dielectric filled.

01:08 So our air filled capacitor, i'm going to call c1.

01:14 And it's going to have a capacitance equal to epsilon knot a divided by d minus t because a t's amount of its thickness is now taken up by the dielectric.

01:29 And then the dielectric capacitor has a capacitance of kappa, epsilon, not a over t.

01:40 And so to find the capacitance of the entire capacitor, we would find the equivalent capacitance.

01:47 And if these are in series, that would be 1 over c1 plus 1 over c2.

01:54 Take the inverse, which would end up being equal to.

01:59 1 .2 times 10 to the negative 10 ferads.

02:07 For part c, we want to figure out what the charge is on the capacitor before you introduce the dielectric.

02:16 So q is just going to be c times v, where again, c is the capacitance we found in part a, and b is the voltage 120 volts...

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