MHTCET 14 May 2023 Shift 1 Solved Paper

To find the electric field at a distance of $0.2\ \mathrm{m}$ from the surface of the conducting sphere, we can use Gauss's Law. For a spherical charge distribution, the electric field at an external point is the same as if all the charge were concentrated at the center of the sphere.First, we calculate the total charge $Q$ on the sphere. The charge density $\sigma$ is the charge per unit area, so multiplying it by the surface area $A$ gives us the total charge $Q$.The surface area of a sphere is given by:$A = 4\pi r^2$ Where $r$ is the radius of the sphere. Plugging in the given radius of $0.1\ \mathrm{m}$,$A = 4\pi (0.1\ \mathrm{m})^2 = 0.04\pi\ \mathrm{m}^2$The charge density $\sigma$ is given as $1.8\ \mathrm{\mu C/m}^2$, which is $1.8 \times 10^{-6}\ \mathrm{C/m}^2$. Now compute the total charge $Q$ :$\; Q = \sigma \cdot A = (1.8 \times 10^{-6}\ \mathrm{C/m}^2) (0.04\pi\ \mathrm{m}^2)$$\; Q = (1.8 \times 10^{-6}) \cdot (0.04\pi)\ \mathrm{C} = 7.2 \times 10^{-8} \pi\ \mathrm{C}$ Gauss's Law states that the electric field $E$ multiplied by the surface area of an imaginary sphere $A'$ that encloses the charge is equal to the total charge $Q$ enclosed divided by the permittivity $\varepsilon_0$ :$E \cdot A' = \frac{Q}{\varepsilon_0}$For a radial distance of $0.2\ \mathrm{m}$ from the surface, the actual radial distance from the center of the sphere will be the sum of the sphere's radius $(0.1\ \mathrm{m}$ ) and the distance from the surface $(0.2\ \mathrm{m}$ ), giving us:$R = 0.1\ \mathrm{m} + 0.2\ \mathrm{m} = 0.3\ \mathrm{m}$ The area of the Gaussian surface $A'$ is now:$A' = 4\pi (0.3\ \mathrm{m})^2 = 0.36\pi\ \mathrm{m}^2$Plugging the values back into the equation we have:$\; E = \frac{7.2 \times 10^{-8} \pi\ \mathrm{C}}{\varepsilon_0 \cdot 0.36 \pi\ \mathrm{m}^2}$$\; E = \frac{7.2 \times 10^{-8}}{0.36 \varepsilon_0}$$\; E = \frac{7.2}{0.36} \times \frac{10^{-8}}{\varepsilon_0}$$\; E = 20 \times \frac{10^{-8}}{\varepsilon_0}\ \mathrm{Vm}^{-1}$$\; E = 2 \times \frac{10^{-7}}{\varepsilon_0}\ \mathrm{Vm}^{-1}$ Now, the value obtained is that the electric field $E$ at a radial distance of $0.2\ \mathrm{m}$ from the sphere's surface is equal to:$E = \frac{2 \times 10^{-7}}{\varepsilon_0}\ \mathrm{Vm}^{-1}$This matches Option C.