Let's solve the problem step by step. ## Problem Recap **Outer sphere radius**: \( R_2 = 20\, \text{cm} = .2\, \text{m} \) (fixed) - **Dielectric**: Air - **Breakdown electric field**: \( E_{max} = 30\, \text{kV/cm} = 3 \times 10^6\, \text{V/m} \) --- ## a) Maximum Voltage, Real and Equivalent Electrode Spacing ### Step 1: **Maximum Electric Field in Spherical Capacitor** For a concentric spherical capacitor, electric field at radius \( r \): \[ E(r) = \frac{V}{r^2} \cdot \frac{r_1 r_2}{r_2 - r_1} \] Alternatively, the potential difference between spheres is: \[ V = \int_{r_1}^{r_2} E(r) dr \] But the maximum electric field occurs at the surface of the **inner sphere** (\( r = r_1 \)): \[ E_{max} = \frac{V}{r_1 (r_2 - r_1)} \] But let's use the correct formula for the field between concentric spheres: \[ E(r) = \frac{V}{r \ln(r_2/r_1)} \] At \( r = r_1 \): \[ E_{max} = \frac{V}{r_1 \ln(r_2/r_1)} \] So, \[ V_{max} = E_{max} \cdot r_1 \ln\left( \frac{r_2}{r_1} \right) \] ### Step 2: **Find Maximum Voltage (Optimized for Breakdown)** To maximize \( V_{max} \), take derivative with respect to \( r_1 \), set to zero, and solve for \( r_1 \): Let \( x = \frac{r_2}{r_1} \): \[ V_{max}(r_1) = E_{max} \cdot r_1 \ln\left( \frac{r_2}{r_1} \right) = E_{max} \cdot r_1 \ln(x) \] But \( r_1 = r_2 / x \), so: \[ V_{max}(x) = E_{max} \cdot \frac{r_2}{x} \ln(x) \] Set derivative to zero: \[ \frac{dV_{max}}{dx} = E_{max} r_2 \cdot \frac{d}{dx} \left( \frac{\ln x}{x} \right ) = \] \[ \frac{d}{dx} \left( \frac{\ln x}{x} \right ) = \frac{1 - \ln x}{x^2} \] Set to zero: \[ 1 - \ln x = \implies \ln x = 1 \implies x = e \] So: \[ \frac{r_2}{r_1} = e \implies r_1 = \frac{r_2}{e} \] ### Step 3: **Calculate Maximum Voltage** Plug \( r_1 = r_2 / e \) into the \( V_{max} \) equation: \[ V_{max} = E_{max} \cdot r_1 \ln \left( \frac{r_2}{r_1} \right ) \] But \( \frac{r_2}{r_1} = e \implies \ln(e) = 1 \) Thus, \[ V_{max} = E_{max} \cdot r_1 \cdot 1 = E_{max} \cdot r_1 \] But \( r_1 = \frac{r_2}{e} = \frac{.2\,\text{m}}{2.718} = .0736\,\text{m} \) So: \[ V_{max} = 3 \times 10^6\,\text{V/m} \cdot .0736\,\text{m} \approx 220,800\,\text{V} = 220.8\,\text{kV} \] ### Step 4: **Calculate Real and Equivalent Electrode Spacing** - **Real spacing:** \( d = r_2 - r_1 = .2 - .0736 = .1264\,\text{m} = 12.64\,\text{cm} \) - **Equivalent spacing:** For a parallel-plate capacitor, \( d_{eq} = V_{max} / E_{max} \) \[ d_{eq} = \frac{V_{max}}{E_{max}} = \frac{220,800}{3 \times 10^6} = .0736\,\text{m} = 7.36\,\text{cm} \] --- ## b) Inner Sphere Radius for 100 kV Now, for \( V = 100\,\text{kV} = 100,000\,\text{V} \): \[ V = E_{max} \cdot r_1 \ln \left( \frac{r_2}{r_1} \right ) \] Let's solve for \( r_1 \): Let \( r_2 = .2\,\text{m} \), \( E_{max} = 3 \times 10^6\,\text{V/m} \), \( V = 100,000\,\text{V} \): \[ 100,000 = 3 \times 10^6 \cdot r_1 \ln \left( \frac{.2}{r_1} \right ) \] \[ \frac{100,000}{3 \times 10^6} = r_1 \ln \left( \frac{.2}{r_1} \right ) \] \[ .03333 = r_1 \ln \left( \frac{.2}{r_1} \right ) \] Let \( y = r_1 \), solve numerically: - Try \( r_1 = .05 \): \( .05 \ln(.2/.05) = .05 \ln(4) = .05 \times 1.386 = .069 \) (too high) - Try \( r_1 = .025 \): \( .025 \ln(.2/.025) = .025 \ln(8) = .025 \times 2.079 = .052 \) (still high) - Try \( r_1 = .01 \): \( .01 \ln(.2/.01) = .01 \ln(20) = .01 \times 3.00 = .03 \) (about right) So, \( r_1 \approx .011\,\text{m} \) to \( .012\,\text{m} \) Let's use a more precise method: \[ r_1 \ln \left( \frac{.2}{r_1} \right ) = .03333 \] Let \( r_1 = x \): Try \( x = .012 \): \( .012 \ln(.2/.012) = .012 \ln(16.67) = .012 \times 2.81 = .0337 \) (very close) So, \[ \boxed{r_1 \approx .012\,\text{m} = 1.2\,\text{cm}} \] Limits: - \( r_1 \) must be **less than** \( r_2 \) (20 cm) - If you want a range, generally \( r_1 > \), but for **100 kV** and maximum breakdown, \( r_1 = 1.2\,\text{cm} \) is the minimum required. --- ## **Summary Table** | Quantity | Value | |-------------------------|------------| | Outer radius \( r_2 \) | 20 cm | | Max field \( E_{max} \) | 30 kV/cm | | Optimal \( r_1 \) | 7.36 cm | | Max voltage \( V_{max} \) | 220.8 kV | | Real spacing \( r_2 - r_1 \) | 12.64 cm | | Equivalent spacing | 7.36 cm | | For 100 kV, \( r_1 \) | 1.2 cm | --- ### **Final Answers** #### a) For optimal breakdown: - **Maximum voltage**: \( \boxed{220.8\,\text{kV}} \) - **Optimal inner sphere radius**: \( \boxed{7.36\,\text{cm}} \) - **Real spacing**: \( \boxed{12.64\,\text{cm}} \) - **Equivalent spacing**: \( \boxed{7.36\,\text{cm}} \) #### b) To apply 100 kV: - **Minimum inner sphere radius**: \( \boxed{1.2\,\text{cm}} \) --- **If you need detailed calculation steps for any part, let me know!**