Parallel plate capacitor and dielectric

In summary: The maximum charge is small, as the capacitance is small.In summary, a parallel-plate capacitor with an area of 1 cm^2 and filled with titanium dioxide (dielectric constant of 80) can have a capacitance of 10^-10 farad with a distance between the plates of 7.08 cm. Its maximum charge, assuming a dielectric strength of 10^8 V/m, would be 7.08*10^-6 coulombs. The relationship between electric field and potential for a constant field is V=Ed, which can be used to find the potential for a capacitor and subsequently its charge.
  • #1
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Homework Statement


a parallel-plate capacitor (2 plates) is filled with titanium dioxide which has a dielectric constant of 80. the area of each plate is 1 cm^2.

a) what must be the spacing between the plates if one wishes to make a 10^-10 farad capacitor?

b) if the dielectric strength of titanium dioxide is 10^8, what is the maximum chrage that this capacitor can safely store?

Homework Equations



dielectric constant kappa = 80
epsilon_0 -- constant = 8.85*10^-12

capacitance C = kappa[(epsilon_0*A)/d] where A is area, given = 1 cm sq, d is distance, unknown.

The Attempt at a Solution



part a:

capacitance C = kappa[(epsilon_0*A)/d] where C = 10^-10 farad, kappa = 80, epsilon_0 = 8.85*10^-12 ---> solve for d, d = 0.0708 meters = 7.08 cm

correct?

part b:

unsure, how does dielectric strength relate to dielectric constant?
what equation should i use, is it similar to one the i used in part a?

thanks
 
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  • #2
Dielectric strength is the maximum electric field that an be applied across the plates before the dielectric breaks down.
 
  • #3
Scholio,

Part a does not look correct to me. (Did you convert the area correctly to m^2? It is not 0.01 m^2.)
 
  • #4
alphysicist, isn't 0.01m = 1cm? do i need to square it?

also i tried part b, assuming dielectric strength(= 10^8 V/m) = max electric field, i used this eq:

electric field E = kq/r^2 where k is constant 9*10^9, r is distance -- from part a =0.0708m, and then solved for q

so q = E*r^2/k = [(10^8)(0.0708^2)]/(9*10^9) = 5.57*10^-5 coulombs

correct?
 
  • #5
scholio said:
alphysicist, isn't 0.01m = 1cm? do i need to square it?

That's right. 1 cm = 0.01m, but what you have here is 1 cm^2, which you might think of as 1 cm cm. A couple of different ways to think the conversion:

[tex]
1 cm^2 \left(\frac{1 m}{100 cm}\right)^2 = 1\times 10^{-4} m^2
[/tex]

or equivalently:

[tex]
1 cm\cdot cm \left(\frac{1 m}{100 cm}\right) \left(\frac{1 m}{100 cm}\right) = 1\times 10^{-4} m^2
[/tex]



also i tried part b, assuming dielectric strength(= 10^8 V/m) = max electric field, i used this eq:

electric field E = kq/r^2 where k is constant 9*10^9, r is distance -- from part a =0.0708m, and then solved for q

so q = E*r^2/k = [(10^8)(0.0708^2)]/(9*10^9) = 5.57*10^-5 coulombs

correct?

A capacitor has its charge spread out over a surface. The charge is not a point charge, and so you cannot use Coulomb's law kq/r^2. For a parallel plate capacitor, we say that the electric field is (approximately) constant. What's the relationship between potential and electric field when the field is constant?
 
  • #6
i changed my area from 0.01 to 1*10^-4m^2 and got a distance, for part a, to be 7.08*10^-4 meters

part b:
as for be, i want to use the eq: capacitance C = epsilon_0*A/d where epsilon_0 = 8.85*10^-12, A = 1*10^-4 m^2, d = 7.08*10^-4m

but the equation is pretty useless if i know all the unknowns. i need an equation with electric field E = dielectric strength because that is given as 10^8 V/m

is potential = electric field when the field is constant? I'm not too sure, couldn't find an equation relating the two.
 
  • #7
I noticed that you had the relationship between field and potential in another thread:

[tex]
\Delta V = - \int\limits_a^b \vec E \cdot d\vec r
[/tex]

What does the right side simplify to when E is a constant field?

Once you have that, how is V and Q related for a capacitor?
 
  • #8
if i did the integral holding E constant would i get r_b - r_a for the right side?

as for the relationship with V and q for a capacitor, capaciance C = Q/V so if i sub in r_b - r_a for V in the C eq, i'd get C = Q/ (r_b - r_a)

correct?
 
  • #9
scholio said:
if i did the integral holding E constant would i get r_b - r_a for the right side?

To be more precise, the E would come out of the integral, and r_b-r_a is the length being integrated over, so the magnitude of the potential would be:

[tex]
\Delta V = E d
[/tex]

(The general formula for a constant electric field is:

[tex]
\Delta V = - E d \cos(\theta)
[/tex]

where [itex]\theta[/itex] is the angle between the field and displacement.)

So you know here that V= Ed, which as you say is used in C=Q/V.


as for the relationship with V and q for a capacitor, capaciance C = Q/V so if i sub in r_b - r_a for V in the C eq, i'd get C = Q/ (r_b - r_a)

correct?
 
  • #10
so since V = Ed, and C = Q/V ---> C = Q/Ed where C = 10^-10 farad, E = 10^8 V/m, d = 7.08*10^-4 meters

so Q = 7.08*10^-6 coulombs, seems a little small

correct?
 
  • #11
That looks right to me.
 

1. What is a parallel plate capacitor?

A parallel plate capacitor is a simple electrical device that is used to store electrical energy by creating an electric field between two parallel conducting plates. It consists of two metal plates separated by a non-conducting material called a dielectric.

2. How does a parallel plate capacitor work?

A parallel plate capacitor works by storing electrical charge on the two plates, creating an electric field between them. The electric field stores energy, which can then be released when the capacitor is connected to a circuit. The capacitance of the capacitor is determined by the distance between the plates, the area of the plates, and the type of dielectric material used.

3. What is the purpose of a dielectric in a parallel plate capacitor?

The dielectric in a parallel plate capacitor serves to increase the capacitance of the device. It does this by reducing the electric field between the plates, allowing for a greater amount of charge to be stored on the plates. Dielectric materials are typically insulators, such as glass, ceramic, or plastic.

4. How does the introduction of a dielectric change the capacitance of a parallel plate capacitor?

The capacitance of a parallel plate capacitor is directly proportional to the permittivity (ε) of the dielectric material used. This means that the higher the permittivity, the greater the capacitance. The distance between the plates and the area of the plates also affect capacitance, but the permittivity of the dielectric has the greatest impact on capacitance.

5. What are some common applications of parallel plate capacitors?

Parallel plate capacitors have a variety of uses, including in electronic circuits, power supplies, and energy storage devices. They are also used in medical equipment, such as X-ray machines, and in radio frequency filters. Additionally, parallel plate capacitors are used in sensors and actuators, and are an important component in the construction of touch screens for smartphones and tablets.

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