JEE Main:

• Cylindrical Capacitor

  • Formula: $$C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$$ where,

    • $$C$$ is the capacitance
    • $$\varepsilon_0$$ is the permittivity of free space $$(\varepsilon_0 = 8.85\times 10^{-12}\text{ C}^2/\text{Nm}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

  • Given values: $$L = 50\text{ cm} = 0.5\text{ m}, \space a = 5\text{ cm} = 0.05\text{ m}, \space b = 10\text{ cm} = 0.1\text{ m}, \space V_{in} = 100\text{ V}, \space V_{out} = 200\text{ V}$$

  • Substituting the values in the formula: $$C = \frac{2\pi\times 8.85\times 10^{-12}\times 0.5}{\ln(0.1/0.05)} = \boxed{2.95\times 10^{-11}\text{ F}}$$

• Spherical Capacitors in Parallel

  • Formula: $$C_{total} = C_1 + C_2$$ where,

    • $$C_{total}$$ is the total capacitance of the parallel combination
    • $$C_1$$ and $$C_2$$ are the capacitances of the individual capacitors

  • Given values: $$R_1 = R_2 = 10\text{ cm} = 0.1\text{ m}, \space d = 20\text{ cm} = 0.2\text{ m}, \space V = 100\text{ V}$$

  • First, we need to calculate the capacitance of each individual capacitor using the formula: $$C = 4\pi\varepsilon_0\frac{R}{d}$$ Substituting the values: $$C_1 = C_2 = 4\pi\times 8.85\times 10^{-12}\times \frac{0.1}{0.2} = 1.77\times 10^{-11}\text{ F}$$

  • Now, we can calculate the total capacitance: $$C_{total} = 1.77\times 10^{-11} + 1.77\times 10^{-11} = \boxed{3.54\times 10^{-11}\text{ F}}$$

• Cylindrical Capacitor

  • Formula: $$C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$$ where,

    • $$C$$ is the capacitance
    • $$\varepsilon_0$$ is the permittivity of free space $$(\varepsilon_0 = 8.85\times 10^{-12}\text{ C}^2/\text{Nm}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

  • Given values: $$L = 10\text{ cm} = 0.1\text{ m}, \space a = 2\text{ cm} = 0.02\text{ m}, \space b = 5\text{ cm} = 0.05\text{ m}, \space V_{in} = 0\text{ V}, \space V_{out} = 100\text{ V}$$

  • Substituting the values in the formula: $$C = \frac{2\pi\times 8.85\times 10^{-12}\times 0.1}{\ln(0.05/0.02)} = \boxed{1.01\times 10^{-10}\text{ F}}$$

CBSE Board:

• Cylindrical Capacitor

  • Formula: $$C = \frac{2\pi\varepsilon_0 L}{\ln(b/a)}$$ where,

    • $$C$$ is the capacitance
    • $$\varepsilon_0$$ is the permittivity of free space $$(\varepsilon_0 = 8.85\times 10^{-12}\text{ C}^2/\text{Nm}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

  • Given values: $$L = 20\text{ cm} = 0.2\text{ m}, \space a = 2\text{ cm} = 0.02\text{ m}, \space b = 4\text{ cm} = 0.04\text{ m}, \space V_{in} = 50\text{ V}, \space V_{out} = 100\text{ V}$$

  • Substituting the values in the formula: $$C = \frac{2\pi\times 8.85\times 10^{-12}\times 0.2}{\ln(0.04/0.02)} = \boxed{1.59\times 10^{-11}\text{ F}}$$

• Spherical Capacitors in Series

  • Formula: $$\frac{1}{C_{total}} = \frac{1}{C_1} + \frac{1}{C_2}$$ where,

    • $$C_{total}$$ is the total capacitance of the series combination
    • $$C_1$$ and $$C_2$$ are the capacitances of the individual capacitors

  • Given values: $$R_1 = R_2 = 5\text{ cm} = 0.05\text{ m}, \space d = 10\text{ cm} = 0.1\text{ m}, \space V = 100\text{ V}$$

  • First, we need to calculate the capacitance of each individual capacitor using the formula: $$C = 4\pi\varepsilon_0\frac{R}{d}$$ Substituting the values: $$C_1 = C_2 = 4\pi\times 8.85\times 10^{-12}\times \frac{0.05}{0.1} = 1.77\times 10^{-11}\text{ F}$$

  • Now, we can calculate the total capacitance: $$\frac{1}{C_{total}} = \frac{1}{1.77\times 10^{-11}} + \frac{1}{1.77\times 10^{-11}}$$ $$\Rightarrow C_{total} = \boxed{0.885\times 10^{-11}\text{ F}}$$

• Spherical Capacitor

  • Formula: $$C = 4\pi\varepsilon_0\frac{R}{d}$$ where,

    • $$C$$ is the capacitance
    • $$\varepsilon_0$$ is the permittivity of free space $$(\varepsilon_0 = 8.85\times 10^{-12}\text{ C}^2/\text{Nm}^2)$$
    • $$R$$ is the radius of the sphere
    • $$d$$ is the distance between the sphere and the outer grounded conductor

  • Given values: $$R = 10\text{ cm} = 0.1\text{ m}, \space d = \infty$$

  • Since the outer conductor is grounded, the distance $$d$$ can be considered infinite. Therefore, the capacitance becomes: $$C = 4\pi\varepsilon_0\frac{R}{\infty} = 0$$

  • Hence, the capacitance of the spherical capacitor is $$\boxed{0\text{ F}}$$.