Given an isolated conducting spherical shell of radius 30 cm. Some positive charges is given to it so that resulting electric field has a maximum intensity of $$1.8×10^6\space\text{NC}^{-1}$$​. The same amount of negative charge is given to another isolated conducting spherical shell of radius 60 cm. Now, first shell is placed inside the second so that both are concentric as given in figure below. Find the electrostatic energy stored in the system?

Correct option is B

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$$\textbf{Given:} \\\bullet \, \text{Maximum electric field at the shell surface: } E_{\text{max}} = 1.8 \times 10^6 \, \text{N/C} \\\bullet \, \text{Radius of the inner shell: } R_1 = 30 \, \text{cm} = 0.3 \, \text{m} \\\bullet \, \text{Radius of the outer shell: } R_2 = 60 \, \text{cm} = 0.6 \, \text{m} \\\bullet \, \text{The charge on the shell: } q = E_{\text{max}} \cdot 4 \pi \epsilon_0 R_1^2 \\\bullet \, \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2 / (\text{N} \cdot \text{m}^2)$$

$$\textbf{Step 1: Finding the charge on the shell:} \\\text{Using the electric field equation at the shell surface:} \\E_{\text{max}} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{q}{R_1^2} \\\text{Rearranging to solve for } q: \\q = E_{\text{max}} \cdot 4 \pi \epsilon_0 R_1^2 \\\text{Substitute the given values:} \\q = (1.8 \times 10^6) \cdot (8.85 \times 10^{-12}) \cdot (0.3)^2 \\q = 1.8 \times 10^{-5} \, \text{C}$$

$$\textbf{Step 2: Finding the capacitance of the system:} \\\text{Now, use the formula for the capacitance of two spherical shells:} \\C = \frac{4 \pi \epsilon_0 R_1 R_2}{R_2 - R_1} \\\text{Substitute the values:} \\C = \frac{4 \pi (8.85 \times 10^{-12}) \cdot (0.3) \cdot (0.6)}{0.6 - 0.3} \\C = \frac{4 \pi \cdot 8.85 \times 10^{-12} \cdot 0.18}{0.3} \\C = \frac{2}{3} \times 10^{-10} \, \text{F} \\\textbf{Step 3: Calculating the electrostatic energy:} \\\text{Now, use the energy formula:} \\U = \frac{Q^2}{2C} \\\text{Substitute the values:} \\U = \frac{(1.8 \times 10^{-5})^2}{2 \times (2/3 \times 10^{-10})} \\U = 2.43 \, \text{J}$$​​​​​