Question

The energy density at a point in a medium of dielectric constant 8 is 26.55 x10⁶ J/m³. Calculate electric field intensity at that point.
(Ans: 0.866 x 10⁹ N/C)

Answer 1

Answer:

0.866 x 10^9 N / C

Explanation:

Given The energy density at a point in a medium of dielectric constant 8 is 26.55 x10⁶ J/m³. Calculate electric field intensity at that point.

Now given u = 26.55 x 10^6 J/m^3 , k = 6

We know that u = 1/2  Ɛo K E^2

                         2 u = Ɛo K E^2

                          E^2 = 2 u /  Ɛo K  

                          E^2 = 2 x 26.55 x 10^6 / 8.85 x 10^-12 x 8

                         E^2 = 53.1 x 10^6 / 70.8 x 10^-12

                         E^2 = 0.75 x 10^18

                        E = 0.866 x 10^9 N / C  

 electric field intensity at that point is 0.866 x 10⁹ N/C