Electrostatics

JEE Main level, single correct, medium

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Question 219 of 311

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

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Answer: (C) 2.25×105 V2.25 \times 10^5 \mathrm{~V} and −5.625×106 V/m-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}

q=1μC=1×10−6C\mathrm{q}=1 \mu \mathrm{C}=1 \times 10^{-6} \mathrm{C}
r=4 cm=4×10−2 m\mathrm{r}=4 \mathrm{~cm}=4 \times 10^{-2} \mathrm{~m}
Potential V=kqr\mathrm{V}=\frac{\mathrm{kq}}{\mathrm{r}}
=9×109×10−64×10−2=2.25×105 V.\begin{aligned} & =\frac{9 \times 10^9 \times 10^{-6}}{4 \times 10^{-2}} \\ & =2.25 \times 10^5 \mathrm{~V} . \end{aligned}
Induced electric field E=−kqr2\mathrm{E}=-\frac{\mathrm{kq}}{\mathrm{r}^2}
=9×109×1×10−616×10−4=−5.625×106 V/m=\frac{9 \times 10^9 \times 1 \times 10^{-6}}{16 \times 10^{-4}}=-5.625 \times 10^6 \mathrm{~V} / \mathrm{m}

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