JEE MAIN - Physics (2023 - 30th January Evening Shift - No. 12)
As shown in the figure, a point charge $Q$ is placed at the centre of conducting spherical shell of inner radius $a$ and outer radius $b$. The electric field due to charge $\mathrm{Q}$ in three different regions $\mathrm{I}, \mathrm{II}$ and $\mathrm{III}$ is given by:
$(\mathrm{I}: r < a, \mathrm{II}: a < r < b$, III: $r>b$ )
_30th_January_Evening_Shift_en_12_1.png)
$(\mathrm{I}: r < a, \mathrm{II}: a < r < b$, III: $r>b$ )
$E_I=0, E_{I I}=0, E_{I I I} \neq 0$
$E_I \neq 0, E_{I I}=0, E_{III}=0$
$E_I \neq 0, E_{I I}=0, E_{III} \neq 0$
$E_I=0, E_{I I}=0, E_{I I I}=0$
Explanation
$${E_I} \ne 0$$ (inside region)
$${E_{II}} = 0$$ (conducting region)
$${E_{III}} \ne 0$$
$$ = {{KQ} \over {{r^2}}}\,\,\,(r > b)$$
Comments (0)
