Two parallel plates have equal and opposite charges. When...

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E=3.20 \times 10^{5} \mathrm{V} / \mathrm{m}$ . When the space is filled with dielectric, the electric field is $E=2.50 \times 10^{5} \mathrm{V} / \mathrm{m}$ . (a) What is the charge density on each surface of the dielectric? (b) What is the dielectric constant?

Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is $E=3.20 \times 10^{5} \mathrm{V} / \mathrm{m}$ . When the space is filled with dielectric, the electric field is $E=2.50 \times 10^{5} \mathrm{V} / \mathrm{m}$ . (a) What is the charge
density on each surface of the dielectric? (b) What is the dielectric constant?

Key Concepts

Dielectric Constant

The dielectric constant, defined as the ratio of the permittivity of a dielectric material to the permittivity of free space (? = ?/??), quantifies how much a dielectric reduces the electric field within a capacitor compared to a vacuum. It is crucial for understanding how the insertion of a dielectric material alters the stored energy and electric field in the capacitor.

Gauss's Law

Gauss's Law connects the electric flux through a closed surface to the total charge enclosed by the surface. In the context of a parallel plate capacitor, it provides a relationship between the surface charge density and the resulting uniform electric field, by using the permittivity of the medium.

Electric Field in Parallel Plate Capacitors

For parallel plate capacitors, the electric field is approximately uniform in the region between the plates. The field is directly related to the surface charge density on the plates and inversely related to the permittivity of the medium present between the plates, using the formula E = ?/?.

Permittivity and Dielectric Materials

Permittivity is a property of a medium that measures how much resistance is encountered when forming an electric field within it. When a dielectric material is introduced between the plates of a capacitor, the effective permittivity increases from ?? (in vacuum) to ? ??, which directly impacts the electric field strength and capacitance.

Transcript

00:01 So in this problem, we have a capacitor and the charges on the capacitor is fixed in this problem.

00:11 And we insert some material later and we found that the electric field before we insert the material is 3 .2 times 10 to the 5th watt meter.

00:27 Let me double check.

00:30 Yeah, fifth.

00:31 And after we insert this dielectric, e prime becomes 2 .5 times 10 to the fifth per meter.

00:42 So in part a, we want to find out the charge density on the dielectric.

00:49 So basically, we know that if, excuse me.

00:54 So for the capacitor, if it is filled with air, we have expression inside the material, between two plates, e equal sigma over epsilon not.

01:07 So where sigma is the charge density on the plate, okay? and you'll see that sigma here equal e times epsilon not, right? so before we insert the material, we have the charge density, sigma equal 2 .83, right? 2 .83 times 10 to the negative 6 column per meter squared.

01:34 But after we insert this material, we have e prime right here, right? so we have sigma prime equal e prime times absolute.

01:45 So this gives us 2 .21 times 10 to the negative 6 column per meter squared.

01:53 So the charge density on the the on the dielectric is simply the difference between sigma and sigma prime.

02:04 So that of sigma equal sigma prime minus sigma.

02:07 So here we choose the absolute value.

02:09 We just want to find out the magnitude...

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