- #1

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How do we prove that ?

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- Thread starter Shreyas Samudra
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- #1

- 166

- 8

How do we prove that ?

- #2

BvU

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- #3

- 166

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yes I know Gauss' Law

but can we prove it

but can we prove it

- #4

cnh1995

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If you have two parallel plates A and B and you put a charge of +10μC on A, it will be distributed on A as +5μC on one surface and +5μC on the other(to make the electric field inside plate A zero). The surface of plate B facing the surface of plate A will have an induced charge of -5μC(and the other surface of plate B will have charge +5μC). This charge is induced in order to terminate the electric field lines from plate A and make the electric field inside plate B zero.yes I know Gauss' Law

but can we prove it

- #5

- 166

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If you have two parallel plates A and B and you put a charge of +10μC on A, it will be distributed on A as +5μC on one surface and +5μC on the other(to make the electric field inside plate A zero). The surface of plate B facing the surface of plate A will have an induced charge of -5μC(and the other surface of plate B will have charge +5μC). This charge is induced in order to terminate the electric field lines from plate A and make the electric field inside plate B zero.

general proof ?

- #6

cnh1995

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Basically, charges are induced on metal plates in order to terminate the external electric field lines and keep electric field inside the plates 0. If the plates are close enough (like in case of capacitors), the charges induced will be "equal and opposite" since the electric field between them is of a constant magnitude σ/2ε.general proof ?

That's all I can say without jumping into math.

- #7

BvU

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Yes. Give it a shot! Where would you locate the Gaussian surface ?yes I know Gauss' Law

but can we prove it

- #8

BvU

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So what would be the flux through the dashed surface ?

- #9

- 166

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yes

thank u

thank u

- #10

BvU

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But if q

- #11

- 166

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we know that the electric field lines are straight , producing zero flux , so that should equal q/ε = 0

hence equal and opposite charges !!

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