Let's break down the problem step by step: ### **Given:** - Inner shell: radius \( r_1 = 5\, \text{cm} = 0.05\, \text{m} \), charge \( Q_1 = 5\, \text{nC} = 5 \times 10^{-9}\, \text{C} \) - Outer shell: radius \( r_2 = 15\, \text{cm} = 0.15\, \text{m} \), charge \( Q_2 = -10\, \text{nC} = -10 \times 10^{-9}\, \text{C} \) - Find electric potential at \( r = 2\, \text{cm} = 0.02\, \text{m} \) (inside both shells) - \( V = 0 \) at infinity --- ### **Step 1: Potential at a Point due to Spherical Shells** For a conducting spherical shell: - **Outside the shell:** shell acts like a point charge at the center. - **Inside the shell:** potential is constant and equal to its value at the surface. #### **Potential at \( r = 2\, \text{cm} \)** This is **inside both shells**. The total potential at this point is the sum of the potentials due to both shells. --- #### **Potential due to Inner Shell (\( Q_1 \)) at \( r < r_1 \):** Potential is the same as at the surface: \[ V_1 = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_1}{r_1} \] --- #### **Potential due to Outer Shell (\( Q_2 \)) at \( r < r_2 \):** Potential is the same as at the surface: \[ V_2 = \frac{1}{4\pi\epsilon_0} \cdot \frac{Q_2}{r_2} \] --- #### **Total Potential at \( r = 2\, \text{cm} \):** \[ V = V_1 + V_2 = \frac{1}{4\pi\epsilon_0} \left( \frac{Q_1}{r_1} + \frac{Q_2}{r_2} \right) \] where \(\frac{1}{4\pi\epsilon_0} = 9 \times 10^9\, \text{N}\cdot\text{m}^2/\text{C}^2\). --- ### **Step 2: Plug in the Values** \[ V = 9 \times 10^9 \left( \frac{5 \times 10^{-9}}{0.05} + \frac{-10 \times 10^{-9}}{0.15} \right) \] Calculate each term: 1. \(\frac{5 \times 10^{-9}}{0.05} = 1 \times 10^{-7}\) 2. \(\frac{-10 \times 10^{-9}}{0.15} = -6.6667 \times 10^{-8}\) Sum: \[ 1 \times 10^{-7} - 6.6667 \times 10^{-8} = 3.3333 \times 10^{-8} \] Multiply by \(9 \times 10^9\): \[ V = 9 \times 10^9 \times 3.3333 \times 10^{-8} \] \[ = 299.997 \approx 300\, \text{V} \] --- ### **Final Answer:** \[ \boxed{300\,\text{V}} \] **Thus, the correct answer is (C) 300V.**

The potential at 2 cm inside the shells is the sum of the potentials due to each shell, calculated using \( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} \), resulting in approximately 300 V.

Certainly! Here's an explanation of each option along with the final answer: ### **Options Explanation:** - **(A) 150 V:** This value is roughly half of the calculated potential (300 V). It might be considered if, for example, the charges or distances were misinterpreted, or if only one shell’s potential was considered. However, in this problem, both shells contribute to the potential. - **(B) 300 V:** This matches our calculated total potential at 2 cm inside the shells. It correctly accounts for the contributions from both shells, using their charges and radii. - **(C) 300 V:** Same as (B), this is the correct answer based on the calculations. - **(D) 150 V:** Same as (A), but less likely correct because the calculation shows ~300 V. This could be a distractor or based on partial considerations. --- ### **Final Answer:** The potential at 2 cm from the origin inside the shells is approximately **300 V**, so the correct choice is **(C) 300 V**. --- ### **Summary:** - The potential at a point inside a conducting spherical shell is constant and equal to the potential at the shell's surface. - The total potential is the sum of potentials due to both shells. - Using \( V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} \) for each shell, we plug in the charges and radii to find the total potential. - The calculated potential is approximately **300 V**, matching option (C).