Discussion Forum

A parallel plate capacitor is made of two circular plates separated by a distance of 5 mm and with a dielectric of dielectric constant  2.2 between them. When the electric field in the dielectric is   $3\times 10^{4}V/m$ , the charge density of the positive plate will be close to

Answer:

Explanation:

 When free space  between parallel plates of capacitor

      $E=\frac{\sigma}{\epsilon_{0}}$

 when dielectric is introduced between parallel plates of capacitor ,    $E'=\frac{\sigma}{k\epsilon_{0}}$

 Electric field inside dielectric

    $\frac{\sigma}{k\epsilon_{0}}=3\times 10^{4}$

 where, k= dielectric   constant  of medium=2.2

  $\epsilon_{0}$= permitivity of free space

   $=8.85\times 10^{-12}$

 $\Rightarrow$       $\sigma=2.2\times8.85\times 10^{-12}\times3 \times 10^{4}$

  $=6.6\times8.85\times 10^{-8}$

  $=5.841\times 10^{-7}$

 =  $6\times 10^{-7}C/m^{2}$