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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 toA. `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)` |
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Answer» Correct Answer - 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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