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## 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?