A positive charge $Q$ is uniformly distributed along a circular ring of radius R.A small test charge $q$ is placed at the centre of the ring figure. Then,
An arbitrary surface encloses a dipole. What is the electric flux through this surface?
From Gauss' law, the electric flux through an enclosed surface is given by $\oint_s E . d S=\frac{q}{\varepsilon_0}$.
Here, $q$ is the net charge inside that enclosed surface.
Now, the net charge on a dipole is given by $-q+q=0$
$\therefore \quad$ Electric flux through a surface enclosing a dipole $=\frac{-q+q}{\varepsilon_0}=\frac{0}{\varepsilon_0}=0$
A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be surface charge density on
(i) the inner surface
(ii) the outer surface?
Here, the charge placed at the centre of the spherical cavity is positively charged. So, the charge created at the inner surface of the sphere, due to induction will be $-Q$ and due to this charge created at outer surface of the sphere is $+Q$.
Now, surface charge density on the inner surface $=\frac{-Q}{4 \pi R_1^2}$
and Surface charge density on the inner surface $=\frac{+Q}{4 \pi R_2^2}$
The dimensions of an atom are of the order of an Angstrom. Thus, there must be large electric fields between the protons and electrons. Why, then is the electrostatic field inside a conductor zero?
The protons and electrons are bound into a atom with distinct and independent existence and neutral in charge. Electrostatic fields are caused by the presence of excess charges. But there can be no excess charge on the inter surface of an isolated conductor. So, the electrostatic fields inside a conductor is zero despite the fact that the dimensions of an atom are of the order of an Angstrom.
If the total charge enclosed by a surface is zero, does it imply that the electric field everywhere on the surface is zero? Conversely, if the electric field everywhere on a surface is zero, does it imply that net charge inside is zero.
Gauss' law also implices that when the surface is so chosen that there are some charges inside and some outside.
The flux in such situation is given by $\oint$ E.dS $=\frac{q}{\varepsilon_0}$.
In such situations, the electric field in the LHS is due to all the charges both inside and outside the surface. The term $q$ on the right side of the equation given by Gauss' law represent only the total charge inside the surface.
Thus, despite being total charge enclosed by a surface zero, it doesn't imply that the electric field everywhere on the surface is zero, the field may be normal to the surface.
Also, conversely if the electric field everywhere on a surface is zero, it doesn't imply that net charge inside it is zero.
i.e., $$\text { Putting } E=0 \text { in } \oint \text { E.dS }=\frac{q}{\varepsilon_0}$$
we get $$q=0.$$