# Finding dielectric constant

• scholio

## Homework Statement

an air gap capacitor is charges with a 50 volt battery. the battery is then disconnected and oil is introduced between the plates. the potential difference between the plate drops 10 volts.

a) what is the dielectric constant?
b) what fraction of the original stored electrical energy is lost when the oil is introduced

## Homework Equations

energy U = 1/2 CV^2 where C is capacitance, V is electric potential

capacitance C = Q/V where Q is charge

capacitance C = kappa[(epsilon_o*A)/d] where kappa = dielectric constant, epsilon_o = 8.85*10^-12, A is area of capacitor, d is distance

## The Attempt at a Solution

part a --> find kappa

use 3rd eq, sub Q/V for C, let Q = 50, V = -10, solve for kappa assuming epsilon_o, A and d are constant to get

kappa = -5d/(epsilon_o*A)

part b.

use first eq, determine energy using capacitance without dielectric constant, then determine energy using capacitance with dielectric constant

do ratio of energy with dielectric/energy no dielectric to determine fraction

correct approach?

scholio said:
part a --> find kappa

use 3rd eq, sub Q/V for C, let Q = 50, V = -10, solve for kappa assuming epsilon_o, A and d are constant to get

kappa = -5d/(epsilon_o*A)

First off, the dielectric constant is the multiple by which capacitance is increased, compared to the capacitance when the capacitor gap is occupied by air (or vacuum, effectively). So kappa is always greater than or equal to 1, and cannot be negative.

Secondly, be careful about the statement of the problem. It doesn't say the new voltage is 10 V -- it says the voltage between the plates has dropped by 10 V. So the new voltage is now...?

With the battery disconnected from the plates, the charge on the plates has nowhere to go. Thus, it will be unchanged by the introduction of the oil dielectric into the gap. We then have

Q = CV = C'V' , where C' is the new capacitance and V' is the new voltage.

Since the original dielectric constant is kappa = 1, the new dielectric constant will be

(kappa)' = kappa · (C'/C) .

part b.

use first eq, determine energy using capacitance without dielectric constant, then determine energy using capacitance with dielectric constant

do ratio of energy with dielectric/energy no dielectric to determine fraction

correct approach?

That will give you the fraction of the original electric potential energy of the capacitor that remains in the "capacitor". Again, be careful: the question asks for the fraction of the original potential energy that has been lost.

It is important to read questions carefully for the requested quantity. Many traps in exam questions (particularly multiple-choice questions) are set in this way...

Incidentally, part (b) can be solved independently of part (a).

BONUS QUESTION: Since the capacitor is not connected to anything, and energy is supposed to be conserved, where did the "lost" electric potential energy go?

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part a:

Q = CV = C' V' let V = 50, V' = 40 so C' = 50C/40 = 5C/4

(kappa)' = kappa(C'/C) where kappa = 1
---> (kappa)' = 1(5C/4C) = 5/4 = 1.25

part b:

U = 1/2(CV^2) = 1/2(C*50^2) = 1250C

U' = 1/2(CV^2) = 1/2((5C/4)*40^2) = 1000C

so U'/U = 1000C/1250C = 1000/1250 = 4/5 fraction

bonus question: is it lost in the air gap?

scholio said:
---> (kappa)' = 1(5C/4C) = 5/4 = 1.25

This should be fine.

so U'/U = 1000C/1250C = 1000/1250 = 4/5 fraction

So the fraction of electric potential energy "lost" is 1/5.

The question can also be answered without using C. The energy stored in the capacitor is also given by U = (1/2)QV . Since the charge is unchanged, we have

U'/U = [ (1/2)QV' ] / [ (1/2)QV ] = V'/V = 40/50 .

bonus question: is it lost in the air gap?

I don't know if your course or textbook goes into the physics of what happens in a dielectric material. The dielectric works the way it does because the electric field between the plates causes a small separation of the charges in the overall neutral dielectric material, referred to as polarization. The field created by the polarization points in the opposite direction to the field between the plates, effectively reducing the strength of the field in the capacitor gap. (The "missing" energy in the capacitor in this problem has gone into producing this charge separation, creating the polarization field.)

If the battery were still connected to the capacitor, this would allow more charge to be stored on the plates for the same level of voltage. You would instead have

V = Q/C = Q'/C' = Q'/[ kappa · C ] ,

so Q' = kappa · Q .

This is the usual reason that dielectric materials are used in capacitors. Storing more charge at the same voltage means more electric potential energy can be stored in the capacitor.

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