A parallel-plate capacitor with plate area A=2.0 m^2 and plate separation d=3.0 mm is connected

A parallel-plate capacitor with plate area A=2.0 m^2 and...

A parallel-plate capacitor with plate area $A=2.0 \mathrm{m}^{2}$ and plate separation $d=3.0 \mathrm{mm}$ is connected to a $45-\mathrm{V}$ battery (Fig. 40 $\mathrm{a} )$ . (a) Determine the charge on the capacitor, the electric field, the capacitance, and the energy stored in the capacitor. (b) With the capacitor still connected to the battery, a slab of plastic with dielectric strength $K=3.2$ is placed between the plates of the capacitor, so that the gap is completely filled with the dielectric. What are the new values of charge, electric field, capacitance, and the energy $U$ stored in the capacitor?

Key Concepts

Parallel-Plate Capacitor

A parallel-plate capacitor consists of two conductive plates separated by a small distance. This configuration creates a uniform electric field between the plates, making it a common system for studying fundamental concepts of capacitance and energy storage in an electric field. Its behavior and properties are derived from the geometry of the plates and the medium in between.

Capacitance

Capacitance is a measure of a capacitor's ability to store electric charge per unit voltage. For a parallel-plate capacitor, capacitance depends on the area of the plates, the separation distance between them, and the properties of the intervening medium. The basic formula without a dielectric is given by C = ??A/d, where ?? is the permittivity of free space, A is the area of the plates, and d is the separation distance.

Electric Field

The electric field in a capacitor is the measure of force per unit charge between the plates and is typically uniform in the region between the plates. It can be determined using the relation E = V/d, where V is the potential difference across the plates, providing insight into how voltage is distributed across the capacitor.

Dielectric Materials

Dielectric materials are insulating substances that, when placed between the plates of a capacitor, increase the capacitor's capacitance by reducing the effective electric field. This is quantified by the dielectric constant of the material. The presence of a dielectric material modifies the original capacitance by a factor equal to the dielectric constant, thereby allowing more charge to be stored for the same applied voltage.

Energy Stored in Capacitor

The energy stored in a capacitor is the work required to separate the charges on its plates and is given by formulas such as U = ½CV², where C is the capacitance and V is the voltage across the plates. This stored energy is a crucial concept in understanding how capacitors can store and release energy in electrical circuits.

Battery Connection and Constant Voltage

When a capacitor is connected to a battery, it is maintained at a constant voltage determined by the battery. This situation means that any change in capacitance—such as through the introduction of a dielectric—will affect the stored charge and energy based on a fixed voltage value, highlighting the relationship between charge, voltage, and capacitance in closed circuits.

Transcript

00:04 This question, you need to use the capacitance formula, a -upsilon 0 divided by d.

00:12 So the first part asks you to find the capacitance.

00:16 So area is given to be 2 meters square.

00:19 So let's substitute that.

00:22 Distance is given 3 millimeters 2 times 8 .8 times 10 power negative 12.

00:38 Divided by distance is 3 millimeters so 3 times 10 power negative negative meters remember your units should be consistent so the answer here is 5 .9 times 10 power negative 9 parallel now the next part is to find the charge on capacitor which is c times v we we are given the potential difference which is 45 so you need to multiply capacitance by 45 you will get the answer as 265 times 5 times 10 power negative 9 cologne since you want to keep the answer in just two significant digits it's 2 .7 times 10 power negative 7 column next part is to find the energy store energy stored in this case is half cv square since we know the c is 5 .9 and applied potential is 45 the answer here is just multiplication of capacitance and potential difference square so your answer will be given by 5973 .7 times 10 power negative 9 so let's keep the answer in just two significant digits so your answer again now is 6 .0 times 10 power negative 6 jule this is energy stored now there is a part which also asks you to find the electric field now electric field let me write it as ef equals v divided by d so the potential difference is 45 distance is 3 millimeters so i need to write in terms of meters this will give me 15 ,000 volt per meter see the units are different for energy is jewel and for electric field it's volt per meter now next part asks you to find the same quantity when the slab of plastic is inserted with the dialectic constant of 3 .2.

04:19 So capacitance now, the final capacitance will be just dielectric constant times the initial capacitance...

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