2. Electric Potential and Capacitance
hard

A point charge of magnitude $+ 1\,\mu C$ is fixed at $(0, 0, 0) $. An isolated uncharged spherical conductor, is fixed with its center at $(4, 0, 0).$ The potential and the induced electric field at the centre of the sphere is

A

$1.8\times 10^5\,V$ and $- 5.625 \times 10^6\,V/m$

B

$0\,V$ and $0\,V/m$

C

$2.25 \times 10^5\,V$ and $-5.625 \times 10^6\,V/m$

D

$2.25 \times 10^5\,V$ and $0\,V/m$

(JEE MAIN-2013)

Solution

$\mathrm{q}=1\, \mu \mathrm{C}=1 \times 10^{-6} \,\mathrm{C}$

$r=4\, \mathrm{cm}=4 \times 10^{-2}\, \mathrm{m}$

$ \text { Potential } V =\frac{\mathrm{k} q}{\mathrm{r}} $

$=\frac{9 \times 10^{9} \times 10^{-6}}{4 \times 10^{-2}} $

$=2.25 \times 10^{5}\, \mathrm{V} $

Induced electric field $\mathrm{E}=-\frac{\mathrm{kq}}{\mathrm{r}^{2}}$

$=\frac{9 \times 10^{9} \times 1 \times 10^{-6}}{16 \times 10^{-4}}=-5.625\, \times 10^{6}\, \mathrm{V} / \mathrm{m}$

Standard 12
Physics

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