What is the radius of the imaginary concentric sphere that divides the electrostatic field of a metal sphere of a radius 20 cm and change of `8muC` in two regions of identical energy?

Options

A. 30 cm
B. 40 cm
C. 60 cm
D. 80 cm

Answer

To find the radius of the imaginary concentric sphere that divides the electrostatic field of a metal sphere of radius 20 cm and charge \(8 \mu C\) into two regions of identical energy, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a metal sphere of radius \(R = 20 \, \text{cm} = 0.2 \, \text{m}\) with a charge \(Q = 8 \, \mu C = 8 \times 10^{-6} \, C\). We need to find the radius \(r\) of a concentric sphere such that the energy in the region between the sphere and this radius is equal to the energy in the region from this radius to infinity.

2. **Define the Energy in Each Region**:
- Let \(U_1\) be the energy in the region from \(R\) to \(r\).
- Let \(U_2\) be the energy in the region from \(r\) to infinity.
- We need to set \(U_1 = U_2\).

3. **Total Energy Calculation**:
- The total energy \(U\) from \(R\) to infinity can be expressed as:
\[ U = U_1 + U_2 \]
- Since \(U_1 = U_2\), we can say:
\[ U = 2U_1 \]

4. **Electric Field of the Sphere**:
- The electric field \(E\) at a distance \(x\) from the center of the sphere (for \(x > R\)) is given by:
\[ E = \frac{kQ}{x^2} \]
- Where \(k\) is Coulomb's constant \(k = \frac{1}{4\pi \epsilon_0}\).

5. **Energy Density**:
- The energy density \(u\) in the electric field is given by:
\[ u = \frac{1}{2} \epsilon E^2 \]

6. **Calculate \(U_1\)**:
- The energy \(U_1\) can be calculated by integrating the energy density over the volume from \(R\) to \(r\):
\[ U_1 = \int_R^r \frac{1}{2} \epsilon E^2 \cdot dV \]
- The differential volume \(dV\) for a spherical shell is \(dV = 4\pi x^2 dx\). Thus:
\[ U_1 = \int_R^r \frac{1}{2} \epsilon \left(\frac{kQ}{x^2}\right)^2 \cdot 4\pi x^2 dx \]

7. **Integrate \(U_1\)**:
- Substitute \(E\) into the integral:
\[ U_1 = \int_R^r \frac{1}{2} \epsilon \frac{k^2 Q^2}{x^4} \cdot 4\pi x^2 dx = 2\pi \epsilon k^2 Q^2 \int_R^r \frac{1}{x^2} dx \]
- The integral evaluates to:
\[ U_1 = 2\pi \epsilon k^2 Q^2 \left[-\frac{1}{x}\right]_R^r = 2\pi \epsilon k^2 Q^2 \left(\frac{1}{R} - \frac{1}{r}\right) \]

8. **Calculate \(U_2\)**:
- The energy \(U_2\) is calculated from \(r\) to infinity:
\[ U_2 = \int_r^\infty \frac{1}{2} \epsilon E^2 \cdot dV = \int_r^\infty \frac{1}{2} \epsilon \left(\frac{kQ}{x^2}\right)^2 \cdot 4\pi x^2 dx \]
- This evaluates to:
\[ U_2 = 2\pi \epsilon k^2 Q^2 \left[-\frac{1}{x}\right]_r^\infty = 2\pi \epsilon k^2 Q^2 \left(0 - \frac{1}{r}\right) = -\frac{2\pi \epsilon k^2 Q^2}{r} \]

9. **Set \(U_1 = U_2\)**:
- Setting the two energies equal gives:
\[ 2\pi \epsilon k^2 Q^2 \left(\frac{1}{R} - \frac{1}{r}\right) = -\frac{2\pi \epsilon k^2 Q^2}{r} \]
- Simplifying leads to:
\[ \frac{1}{R} - \frac{1}{r} = -\frac{1}{2r} \]
- Rearranging gives:
\[ \frac{1}{R} = \frac{1}{2r} + \frac{1}{r} = \frac{3}{2r} \]
- Thus:
\[ r = \frac{3}{2} R \]

10. **Calculate the Final Radius**:
- Substituting \(R = 20 \, \text{cm}\):
\[ r = \frac{3}{2} \times 20 \, \text{cm} = 30 \, \text{cm} \]

### Final Answer:
The radius of the imaginary concentric sphere is **30 cm**.