Consider an atom with atomic number Z as cons | vedclass

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(A) The total electric field $E$ at a distance $r$ $(r < R)$ is the vector sum of the field due to the central positive charge and the field due to th

Vedclass · Accepted Answer

$\frac{Ze}{4\pi\varepsilon_0}\left[\frac{1}{r^2}-\frac{r}{R^3}\right]$

Vedclass · Answer

(A) The total electric field $E$ at a distance $r$ $(r $1$. The electric field due to the central positive charge $+Ze$ at distance $r$ is $E_1 = \frac{1}{4\pi\varepsilon_0} \frac{Ze}{r^2}$.$2$. The negative charge $-Ze$ is uniformly distributed in a sphere of radius $R$. Using Gauss's Law,the electric field $E_2$ due to this sphere at distance $r$ is $E_2 = \frac{1}{4\pi\varepsilon_0} \frac{(Ze)r}{R^3}$.$3$. Since the fields are in opposite directions,the net electric field is $E = E_1 - E_2 = \frac{Ze}{4\pi\varepsilon_0} \left[ \frac{1}{r^2} - \frac{r}{R^3} \right]$.

Consider an atom with atomic number $Z$ as consisting of a positive point charge at the centre and surrounded by a distribution of negative electricity uniformly distributed within a sphere of radius $R$. The electric field at a point inside the atom at a distance $r$ from the centre is

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