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The sketech below show cross secion so e...

The sketech below show cross secion so equipotential surfaces between two charged.Conductors that are shwon in solid black.Some points on the wquipotenital surfaces .near the conductors are marked as `A,B,C…. .` The arrangements lies in air (Take `epsilon_(0)=8.85xx10^(-12)C^(2)//Nm^(2)]`

Surfaces charge density of the plate is equal to

A

`8.85xx10^(-10)C//m^(2)`

B

`-8.85xx10^(-10)C//m^(2)`

C

`17.7xx10^(-10)C//m^(-2)`

D

`-17.7xx10^(-10)C//m(2)`

Text Solution

Verified by Experts

The correct Answer is:
A
`E=(40-10)/(0.3)=100V//m`
(near the plate the electric field has to be uniform \it is alomost due to the particle
For conducting plate
`E=(sigma)/(epsilon_(0))sigma=epsilon_(0)E`
therefore `s=8.85xx10^(-12)xx100`
`=8.85xx10^(-10)C//m^(2)`
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