### Given Data: * Outer sphere radius: \( b = 20 \, \text{cm} \) (fixed) * Inner sphere radius: \( a \) (unknown) * Dielectric medium: air * Breakdown field strength: \( E_{\text{bd}} = 30 \, \text{kV/cm} \) We have a concentric spherical electrode system. --- ### Step 1. Basic relations For a spherical capacitor: \[ V = \frac{Q}{4\pi \varepsilon_} \left( \frac{1}{a} - \frac{1}{b} \right) \] and \[ E(r) = \frac{Q}{4\pi \varepsilon_ r^2}. \] The **maximum field** occurs at the **inner sphere surface** (\( r = a \)): \[ E_{\max} = E(a) = \frac{Q}{4\pi \varepsilon_ a^2}. \] We can express \( Q/(4\pi\varepsilon_) \) in terms of \( V \): \[ \frac{Q}{4\pi\varepsilon_} = \frac{V}{\frac{1}{a} - \frac{1}{b}} = V \frac{ab}{b - a}. \] Substitute into \( E_{\max} \): \[ E_{\max} = V \frac{b}{a(b - a)}. \] At the breakdown limit (\( E_{\max} = E_{\text{bd}} \)), the **maximum allowable voltage** is: \[ \boxed{V_{\max} = E_{\text{bd}} \frac{a(b - a)}{b}}. \] --- ## (a) To find the optimum configuration (maximum possible \(V_{\max}\)) We want to **maximize** \(V_{\max}(a) = E_{\text{bd}} \frac{a(b - a)}{b}\). Differentiate with respect to \(a\): \[ \frac{dV_{\max}}{da} = E_{\text{bd}} \frac{1 - 2a/b}{1} = \quad\Rightarrow\quad a = \frac{b}{2}. \] So the maximum voltage occurs when the **inner radius is half** of the outer radius. * Given \( b = 20\,\text{cm} \), we get \( a = 10\,\text{cm}. \) Now calculate: \[ V_{\max} = E_{\text{bd}} \frac{a(b - a)}{b} = 30\,\text{kV/cm} \times \frac{10(20 - 10)}{20} = 30 \times 5 = 150\,\text{kV}. \] ✅ **Maximum voltage before breakdown:** \[ \boxed{V_{\max} = 150\,\text{kV}} \] ✅ **Actual electrode spacing:** \[ b - a = 20 - 10 = 10\,\text{cm}. \] ✅ **Equivalent flat-plate spacing:** The equivalent gap (\(d_{\text{eq}}\)) is found from \( V_{\max} = E_{\text{bd}} d_{\text{eq}} \): \[ d_{\text{eq}} = \frac{V_{\max}}{E_{\text{bd}}} = \frac{150}{30} = 5\,\text{cm}. \] So, even though the physical gap is 10 cm, it behaves like a 5 cm flat gap in terms of breakdown voltage. --- ## (b) To apply **100 kV** safely, find the allowable range of \(a\) Condition for no breakdown: \[ E_{\max} = V \frac{b}{a(b - a)} \le E_{\text{bd}}. \] Rearrange: \[ a(b - a) \ge \frac{V b}{E_{\text{bd}}}. \] Compute the constant: \[ c = \frac{V b}{E_{\text{bd}}} = \frac{100 \times 20}{30} = 66.67\,\text{cm}^2. \] So: \[ a(b - a) \ge 66.67. \] At the limit (equality): \[ a(b - a) = c \Rightarrow a^2 - b a + c = . \] Solve for \(a\): \[ a = \frac{b \pm \sqrt{b^2 - 4c}}{2}. \] Substitute \(b = 20\) cm and \(c = 66.67\): \[ a = \frac{20 \pm \sqrt{400 - 266.67}}{2} = \frac{20 \pm \sqrt{133.33}}{2}. \] \[ \sqrt{133.33} \approx 11.55. \] Thus: \[ a_1 = \frac{20 - 11.55}{2} = 4.23\,\text{cm}, \quad a_2 = \frac{20 + 11.55}{2} = 15.77\,\text{cm}. \] Because \(V_{\max}\) is maximum at \(a = 10\) cm (center of the parabola), the field will be below breakdown for values between these limits. ✅ **Hence, for 100 kV operation:** \[ \boxed{4.23\,\text{cm} \le a \le 15.77\,\text{cm}} \] --- ### ✅ Final Answers | Quantity | Symbol | Result | | ------------------------ | ---------------- | ---------------------- | | Outer radius | \(b\) | 20 cm | | Optimum inner radius | \(a_{\text{opt}}\) | 10 cm | | Maximum safe voltage | \(V_{\max}\) | 150 kV | | Actual electrode spacing | \(b - a\) | 10 cm | | Equivalent flat gap | \(d_{\text{eq}}\) | 5 cm | | Allowable \(a\) for 100 kV | — | 4.23 cm ≤ a ≤ 15.77 cm |