Physics M.C.Q [1M]

1. The flux of the electric field through the sphere is zero.

3. The electric field is not zero anywhere on the sphere.

Explanation:

$\therefore$ Net charge of electric dipole is always zero.

$\phi=\frac{\text{q}_{\text{inside}}}{\in_0}$ Here inside = 0

The flux of the electric field through the sphere is zero. But the electric field is not zero any where on the sphere. Take 'A' point any where on the sphere, and find the electric field on the surface. You will get non zero electric field on the sphere.

2. Will become positive.

Explanation:

Charge distribution:

Due to charge distribution, intersection by the closed surface of conducting sphere containing the positive charge. So we can say that the flux of the electric field through the closed surface will become positive.

$\therefore\phi=\frac{\text{q}_{\text{in}}}{\in_0}\Rightarrow+\text{ve}$

2. Towards right.

Explanation:

Force on the central charge due to shell is right $\frac{\text{Kq}^2}{\text{r}^2}$ towards.

2. The electric field may be zero everywhere in the surface.
3. The charge inside the surface must be zero.

Explanation:

$\phi=\frac{\text{q}_{\text{inside}}}{\in_0}=\int{\text{E}}.\text{d}{\text{S}}$

If the flux of the electric field through a closed surface is zero mean the net charge inside the surface must be zero or the electric field may be zero every where on the surface.

3. $\frac{\text{q}}{2\in_0}$

Explanation:

In the charge 'q' in side closed cylinder, due to charge the flux of the electric field through the surface of the vessel is $\frac{\text{q}}{\in_0}.$ Here A charge q is placed at the centre of the open end of a cylindrical, By symmetricily Half line of flux goes outside & half flux line goes inside. So we can say that the flux of the electric field through the surface of the vessel is $\frac{\text{q}}{2\in_0}.$