Question

calculate the value of electric field intensity at the surface of conducter ​

Answer 1

Answer:

The electric field is zero inside a conductor.  

Explanation:

In electrostatics free charges in a good conductor reside only on the surface. So the free charge inside the conductor is zero.

Answer 2

Answer:

The value of electric field intensity at the surface of conductor is $\frac{\sigma}{\epsilon_0}$ .

Explanation:

According to the Gauss's law,

              $\int \vec{E} . \vec{ds} = \frac{q}{\epsilon_0}$      [E → Electric Field, q → total charge]

       →  $\int {E} {ds} \cos0^o= \frac{q}{\epsilon_0}$

      →  $E \int ds = \frac{q}{\epsilon_0}$ .......................................................(i)

Now,$q= \int \sigma ds = \sigma \int ds$       [ $\sigma$ → uniform surface charge density]

Putting the value of 'q' in equation(i),we get

                                  $E \int ds = \frac{q}{\epsilon_0} = \frac{\sigma}{\epsilon_0} \int ds$

                                →   $E = \frac{\sigma}{\epsilon_o}$

∴The value of electric field intensity at the surface of conductor is $\frac{\sigma}{\epsilon_0}$ .