Capacitor Exercise (09-08-2020)

The gap between the plates of a parallel plate capacitor is filled with isotropic dielectric whose permittivity ϵ \epsilon varies linearly from ϵ 1 \epsilon_{1} to ϵ 2 \epsilon_{2} ( ϵ 2 > ϵ 1 ) (\epsilon_{2}> \epsilon_{1} ) in the direction perpendicular to the plates.
The area of each plates equals S S , the separation between the plates is equal to d d .
Find the space density of the bound charges as a function of ϵ \epsilon if the charge of the capacitor is q and the field E E
in it is directing toward the growing ϵ \epsilon value.

The answers comes in the form of α q ( ϵ 2 ϵ 1 ) β S d ϵ γ \frac{\alpha q(\epsilon_{2}- \epsilon_{1})^{\beta} }{Sd \epsilon^{\gamma}}
Then type your answer as α + β + γ = ? \alpha+\beta+\gamma=?


The answer is 2.

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1 solution

Here α = 1 , β = 1 , γ = 2 α=-1,β=1,\gamma =2

So α + β + γ = 2 α+β+\gamma =\boxed 2

@Foolish Learner please post the solution of ring and magnetism.

Talulah Riley - 10 months, 1 week ago

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I think the answer should be m g 2 π B 0 R \dfrac {mg}{2πB_0R} .

A Former Brilliant Member - 10 months, 1 week ago

@Foolish Learner The most smallest solution I have ever seen . . You should post a method rather than just telling α , β , γ \alpha, \beta, \gamma values??

Talulah Riley - 10 months ago

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