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?

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

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.

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.

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.

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.

Transcript

00:01 In this problem on the topic of capacitance, we have a parallel plate capacitor, which have plates of an area of 0 .1 -2 square meters and separation 1 .2 centimeters.

00:10 A battery charges the plates to a potential difference of 120 volts before being disconnected.

00:17 And we then place a dielectric slab, which has a dielectric constant of 4 .8 and thickness 4 millimeters symmetrically between the plates.

00:25 We want to find the capacitance before the slab is inserted.

00:28 The capacitance with a slab in place, the free charge before and after the slab is inserted, the magnitude of the electric field, firstly in the space between the plates, and then between the plates and dielectric, and then in the dilectric itself, the potential difference across the plates with a slab in place, and lastly, the external work that is involved to insert the slab.

00:51 Now, initially, the capacitance is c -0, which is epsilon -not times the area a over the plate separation d, which is the constant epsilon not, which is 8 .85 times 10 to the minus 12, coulum squared per newton meter squared, times the area 0 .12 square meters divided by 1 .2 times 10 to the minus 2 meters, which gives the initial capacitance to be 89 picofarons.

01:34 Now for part b, we find the capacitance c to be epsilon 0 times a times kappa over kappa into d minus b plus b, putting in our values, this is 8 .85 times 10 to the minus 12 in sir units, which are suppressed, times the area of 0 .12 square meters.

02:11 Times the dielectric constant 4 .8 divided by the dielectric constant kappa 4 .8 into d minus b, which is 1 .2, minus 0 .4 times 10 to the minus 2 meters, all of this plus b, which is 4 times 10 to the minus 3 meters.

02:42 And so calculating we get this capacitance c to be 1 .2 times 10 to the power to picofarons.

02:59 Now, for part c, before the insertion, we have the charge q to be c -0 times v, which is 89 picofarad times 120 volts, which gives the charge of 11 nanoculums...

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