Electric Charges and Fields Test - 20

When a positive point charge +q is placed near an isolated conducting plane, some negative charge developes on the surface of the plane towards the charge and an equal positive charge developes on opposite side of the plane. This is called induction process and the electric field on a isolated conducting plane at point is directly projected in a plane perpendicular to the field and away from the plane.

Consider a point on diameter away from the centre of hemisphere which is uniformly positively charged, then the electric field is perpendicular to the diameter and the component of electric intensity parallel to the diameter cancel out.

In general Gauss's law is applied to any closed surface.

In Gauss's law, $\oint\limits_s E.dS$ means the algebraic sum of number of flux lines entering the surface and number of flux lines leaving the surface.

When $\oint\limits_s E.dS=0$, the number of flux lines entering the surface must be equal to the number of flux lines leaving it, if there is no change inside.

$\oint\limits_s E.dS={q\over \varepsilon_0}$, where q is charge enclosed by the surface, and $\in_0$ is permitivity.

If $\oint\limits_s E.dS=0$, q = 0 then the resultant charge enclosed by the surface must be zero.

So, all other charges must necessarily be outside the surface. This is because of the fact that charges outside the surface do not contribute to the electric flux.

Due to a charge Q the electric field at a point in space may be defined as the electric force per unit charge on a charged object at that point. Thus, electric field due to the charge Q will be continuous, if there is no charge at that point. It will be discontinuous if there is a charge at that point.

By Gauss’ law, $\oint\limits_sE.dS={q\over \varepsilon_0}$, where q is the net charge enclosed within the surface. If $\oint\limits_sE.dS=0$, then q = 0, i.e., net charge enclosed by the surface must be zero. So all the charge is outside the surface. Here, electric flux doesn’t depend on the type or nature of charge.

The electric field due to dipole is proportional to $({1\over r^3})$. When there are various types of charges in a region, but the net charge is zero, the region is, supposed to contain a number of electric dipoles.

So, at points outside the region the dominant electric field $\propto {1\over r^3}$ for large r.

Then, electric field is conservative, work done to move a charged particle along a closed path, away from the region will be zero.