A parallel plate capacitor used in a flash for a camera must...

A parallel plate capacitor used in a flash for a camera must be able to store 32 J of energy when connected to $300 \mathrm{V} .$ (Most electronic flashes actually use a $1.5 \mathrm{to}$ $6.0 \mathrm{V}$ battery, but increase the effective voltage using a dc-dc inverter.) (a) What should be the capacitance of this capacitor? (b) If this capacitor has an area of $9.0 \mathrm{m}^{2}$ and a distance between the plates of $1.1 \times 10^{-6} \mathrm{m},$ what is the dielectric constant of the material between the plates? (The large effective area can be put into a small volume by rolling the capacitor tightly in a cylinder.) (c) Assuming the capacitor completely discharges to produce a flash in $4.0 \times 10^{-3}$ s, what average power is dissipated in the flashbulb during this time? (d) If the distance between the plates of the capacitor could be reduced to half its value, how much energy would the capacitor store if charged to the same voltage?

A parallel plate capacitor used in a flash for a camera must be able to store 32 J of energy when connected to $300 \mathrm{V} .$ (Most electronic flashes actually use a $1.5 \mathrm{to}$ $6.0 \mathrm{V}$ battery, but increase the effective voltage using a dc-dc inverter.) (a) What should be the capacitance of this capacitor? (b) If this capacitor has an area of $9.0 \mathrm{m}^{2}$ and a distance between the plates of $1.1 \times 10^{-6} \mathrm{m},$ what is the dielectric constant of the material between the plates? (The large effective area can be put into a small volume by rolling the capacitor tightly in a cylinder.)
(c) Assuming the capacitor completely discharges to produce a flash in $4.0 \times 10^{-3}$ s, what average power is dissipated in the flashbulb during this time?
(d) If the distance between the plates of the capacitor could be reduced to half its value, how much energy would the capacitor store if charged to the same voltage?

Key Concepts

Parallel Plate Capacitors

Parallel plate capacitors are a basic type of capacitor characterized by two conductive plates separated by a distance, with an insulating medium between them. Their capacitance is directly proportional to the area of the plates and the dielectric constant of the medium, and inversely proportional to the distance between the plates. This concept helps in analyzing how physical design parameters influence the overall performance of the capacitor.

Capacitor Energy Storage

This concept focuses on the ability of a capacitor to store energy as an electric field. The energy stored is given by the formula E = 1/2 C V², where C is the capacitance and V is the voltage. This relationship is fundamental in understanding how different capacitor configurations and operating conditions affect the amount of energy that can be stored and later released.

Dielectric Materials and Dielectric Constant

Dielectric materials are insulators used between the plates of a capacitor to increase its capacitance without allowing current to flow. The dielectric constant (or relative permittivity) is a measure of how much the material can increase the capacitance compared to a vacuum. Understanding this concept is crucial for designing capacitors that efficiently store energy in a given physical space.

Electrical Power Dissipation during Discharge

When a capacitor discharges, the energy stored is rapidly released, which can be quantified as power dissipation. The average power is calculated by dividing the total energy by the time over which the discharge occurs. This concept is important in applications such as camera flashes, where a quick, intense burst of energy is required.

Effect of Capacitor Geometry on Energy Storage

The geometry of a capacitor, including the area of the plates and the distance between them, plays a critical role in determining its capacitance and hence the energy stored. Modifying geometrical parameters, such as reducing the plate separation, directly impacts the capacitance and the energy storage capacity. Understanding this relationship is key in designing capacitors for specific energy and size requirements.

Transcript

0:00 Hi.

00:01 In the given problem, energy stored in the parallel plate capacitor which is being used in flash of a camera is given as u is equal to 32 joules.

00:40 And the potential applied to it is 300 volt.

00:46 Now you in the first part of the problem using the expression.

00:51 For the energy stored in a capacitor which is given as u is equal to half c v square the expression for the unknown capacitance will come out to be two of u divided by v square so plugging in all known values this is two times of energy which is 32 divided by square of potential which is 300 volt so this capacitance here comes out to be 7 .1 into 10 to the power minus 4 fared which is answered for the first part of this problem.

01:29 Now in the second part of the problem, area of the plates of this parallel plate capacitor is given as 9 .0 meter square and the gap between the plates is d is equal to 1 .1 into 10 to the power minus 6 meter.

01:50 So using the expression for the capacitance of a parallel plate capacitor having a dielectric between its plates, this is given as c is equal to epsilon not a, k by d, where k is the dialectic constant of that medium.

02:05 So an expression for the dialectic constant here will come out to be c multiplied by d divided by epsilon not into so again plugging in all known values for a c this is 7 .1 into 10 dash power minus 4 which we have obtained in the first part then d this is given as 1 .1 into 10 dash per minus 6 meter for epsilon node this is 8 .854 into 10 tish par minus 12 and finally for area this is 9 .0 meter is square so so this dialectic constant then is calculated to be 9 .8.

02:49 Answer for the second part of this problem...

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