Terminology of electric flux density

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Electric flux density D is defined as εE in linear and isotropic media, where E is the electric field and ε is permittivity. A linear medium means the response to the electric field is proportional, while an isotropic medium behaves uniformly in all directions. Understanding these terms is essential for applying the concept correctly in physics. The discussion also highlights a preference for the term "electric displacement field" over "electric flux density." Clarity on these definitions is crucial for accurate interpretation in electromagnetic theory.
Miike012
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In my book is says the electric flux density D is equal to εE if the medium is linear and isotropic, where E is the electric field and permittivity ε is a scalar.

I have no idea what they mean by a linear and isotropic medium.. How am I suppose to know if the medium is linear or isotropic?
 
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Have you tried looking the words up in a dictionary or elsewhere?
i.e. http://en.wikipedia.org/wiki/Permittivity

For instance "isotropic" just means that the medium behaves much the same in every direction ...
i.e. the speed of light is the same in every direction.
"linear" means that if electric field of strength E has effect size A, then an electric field of strength 2E has effect 2A.

aside: I'd usually refer to D as the "electric displacement field" rather than the "electric flux density".
http://en.wikipedia.org/wiki/Electric_displacement_field
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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