Question

Two metal spheres having charge densities $5\mu C/m^2$ and $-2\mu C/m^2$ with radii $2mm$ and $1mm$ respectively are kept in a hypothetical closed surface. Calculate total normal electric induction over the closed surface.

Answer 1

Given, ${\sigma }_1=5\mu C/m^2$

${\sigma }_2=-2\mu C/m^2$

$r_1=2mm$, $r_2=1mm$.

Charge enclosed by each sphere,

$q_1=4\pi r^2\times {\sigma }_1$

$q_1=4\times 3.14\times {(2\times {10}^{-3})}^2\times 5\times {10}^{-6}$

$q_1=4\times 3.14\times 20\times {10}^{-12}$

$q_1=251.2\times {10}^{-12}C$

$q_2=4\pi r^2\times {\sigma }_1$

$=4\times 3.14\times {\left({10}^{-3}\right)}^2\times (-2\times {10}^{-6})$

$=-12.56\times 2\times {10}^{-12}$

$q_2=-25.12\times {10}^{-12}C$

According to the T.N.E.I. (Total Normal Electric Induction), "it is equal to the total charge enclosed by that surface".

T.N.E.I. $=q$

$=q_1+q_2$

$=251.2\times {10}^{-12}+(-25.12\times {10}^{-12})$

$=226.08\times {10}^{-12}N{m}^2/C$