Question

solve this :----------​

Answer 1

Proof:-

Ф = -∫EdA (By Definition)

Here Ф is the flux

E is the Electric field.

dA is a small area taken.

Ф = -EA (Integration of dA is A)

Now, formula of electric field is given by:

$E=\frac{q}{4\pi (eo)r^{2} }$

Here, eo is the permitivity.

Ф= $E=\frac{q}{4\pi (eo)r^{2} }*4\pi r^{2}$

(Area of a sphere is 4$\pi r^{2}$)

(Here we take the closed surface as sphere)

Thus, we get :

Ф $=\frac{q}{eo}$

THIS THEOREM IS KNOWN AS GAUSS THEOREM.

If the charge lies outside the closed surface, the electric flux will 0 as Ф=$\frac{q}{eo}$ where q is the charge enclosed (Inside the surface)