# Parallel plate capacitor and dielectric

## 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

rl.bhat
Homework Helper
Dielectric strength is the maximum electric field that an be applied across the plates before the dielectric breaks down.

alphysicist
Homework Helper
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.)

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?

alphysicist
Homework Helper
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:

$$1 cm^2 \left(\frac{1 m}{100 cm}\right)^2 = 1\times 10^{-4} m^2$$

or equivalently:

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

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?

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.

alphysicist
Homework Helper
I noticed that you had the relationship between field and potential in another thread:

$$\Delta V = - \int\limits_a^b \vec E \cdot d\vec r$$

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?

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?

alphysicist
Homework Helper
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:

$$\Delta V = E d$$

(The general formula for a constant electric field is:

$$\Delta V = - E d \cos(\theta)$$

where $\theta$ 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?

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?

alphysicist
Homework Helper
That looks right to me.