- #26

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Oh okay, well ifI asked the relation between the new and the initial surface charges.

σ=ε

_{0}E

_{1}

σ'=0.7σ=0.7ε

_{0}E

_{1}

Or would it be by a factor of 0.3?

- Thread starter Callix
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- #26

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Oh okay, well ifI asked the relation between the new and the initial surface charges.

σ=ε

σ'=0.7σ=0.7ε

Or would it be by a factor of 0.3?

- #27

ehild

Homework Helper

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What do you think, is the new charge more or less than the initial one? Why we use dielectrics in the capacitors? To increase their ability to store more charge.Oh okay, well if

σ=ε_{0}E_{1}

σ'=0.7σ=0.7ε_{0}E_{1}

Or would it be by a factor of 0.3?

Look at your picture. The actual E is E = E1-E2, and E2=0.3 E.

- #28

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Right, so then it would be σ'=0.7σ=0.7εWhat do you think, is the new charge more or less than the initial one? Why we use dielectrics in the capacitors? To increase their ability to store more charge.

Look at your picture. The actual E is E = E1-E2, and E2=0.3 E.

- #29

ehild

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Is the new charge less than the initial one?

- #30

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Well logically I would suppose not then, since C=q/V and we already know V is constant because of the battery, so q must change. Adding dielectrics help increase C, so that means q increases as well.Is the new charge less than the initial one?

So before, were you saying that E=E1-(-E2)? Or E=E1+(-E2), because it seems like you're hinting that it's the first, which in that case it would be by a factor of 1.7.

- #31

ehild

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It is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.Well logically I would suppose not then, since C=q/V and we already know V is constant because of the battery, so q must change. Adding dielectrics help increase C, so that means q increases as well.

So before, were you saying that E=E1-(-E2)? Or E=E1+(-E2), because it seems like you're hinting that it's the first, which in that case it would be by a factor of 1.7.

- #32

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Right, so then it would increase by a factor of 1.7, soIt is the first, E1=E+E2, but E2 is the field of the dipoles, which is 0.3E, if the E2 means the magnitude of the dipole field.

σ'=1.7σ=1.7ε

And that is the surface charge density on the dielectric?

- #33

ehild

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Why 1.7? It is wrong.Right, so then it would increase by a factor of 1.7, so

σ'=1.7σ=1.7ε_{0}E_{1}.

And that is the surface charge density on the dielectric?

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