NEET 2020 Physics Question 151 Solution

For a spherical conductor of radius $$R$$ carrying total charge $$Q$$ uniformly distributed on its surface: - Inside the conductor ($$r < R$$): $$E = 0$$. - Outside the conductor ($$r \ge R$$): By spherical symmetry and Gauss's law, the electric field is identical to that of a point charge $$Q$$ located at the center: $$E = \frac{1}{4\pi\varepsilon_0} \frac{Q}{r^2}$$

The point is at a distance $$r = 15\text{ cm}$$ from the centre. Because $$r = 15\text{ cm} > R = 10\text{ cm}$$, the point lies outside the sphere. $$r = 15\text{ cm} = 0.15\text{ m} = \frac{3}{20}\text{ m}$$

$$E = \frac{(9 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (3.2 \times 10^{-7}\text{ C})}{(0.15\text{ m})^2}$$ $$E = \frac{28.8 \times 10^2}{0.0225} = \frac{2880}{0.0225}$$ Multiply numerator and denominator by 400/9: $$E = 2880 \times \frac{400}{9} = 320 \times 400 = 128000\text{ N C}^{-1} = 1.28 \times 10^5\text{ N C}^{-1}$$

Comparing with options, $$1.28 \times 10^5\text{ N/C}$$ is uniquely Option (2).