JEE Main:

• Cylindrical Capacitor

  • Formula: $$C=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}$$ where,

    • $$C$$ is the capacitance
    • $${\epsilon }_0$$ is the permittivity of free space $$({\epsilon }_0=8.85\times {10}^{-12}{\text{ C}}^2/{\text{Nm}}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

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

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

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

  • First, we need to calculate the capacitance of each individual capacitor using the formula: $$C=4\pi {\epsilon }_0\frac{R}{d}$$ 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: $$C_{total}=1.77\times {10}^{-11}+1.77\times {10}^{-11}=\boxed{{\displaystyle 3.54\times {10}^{-11}\text{ F}}}$$

• Cylindrical Capacitor

  • Formula: $$C=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}$$ where,

    • $$C$$ is the capacitance
    • $${\epsilon }_0$$ is the permittivity of free space $$({\epsilon }_0=8.85\times {10}^{-12}{\text{ C}}^2/{\text{Nm}}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

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

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

CBSE Board:

• Cylindrical Capacitor

  • Formula: $$C=\frac{2\pi {\epsilon }_{0}L}{\ln (b/a)}$$ where,

    • $$C$$ is the capacitance
    • $${\epsilon }_0$$ is the permittivity of free space $$({\epsilon }_0=8.85\times {10}^{-12}{\text{ C}}^2/{\text{Nm}}^2)$$
    • $$L$$ is the length of the capacitor
    • $$a$$ is the radius of the inner cylinder
    • $$b$$ is the radius of the outer cylinder

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

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

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

  • First, we need to calculate the capacitance of each individual capacitor using the formula: $$C=4\pi {\epsilon }_0\frac{R}{d}$$ 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: $$\frac{1}{C_{total}}=\frac{1}{1.77\times {10}^{-11}}+\frac{1}{1.77\times {10}^{-11}}$$ $$\Rightarrow C_{total}=\boxed{{\displaystyle 0.885\times {10}^{-11}\text{ F}}}$$

• Spherical Capacitor

  • Formula: $$C=4\pi {\epsilon }_0\frac{R}{d}$$ where,

    • $$C$$ is the capacitance
    • $${\epsilon }_0$$ is the permittivity of free space $$({\epsilon }_0=8.85\times {10}^{-12}{\text{ C}}^2/{\text{Nm}}^2)$$
    • $$R$$ is the radius of the sphere
    • $$d$$ is the distance between the sphere and the outer grounded conductor

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

  • Since the outer conductor is grounded, the distance $$d$$ can be considered infinite. Therefore, the capacitance becomes: $$C=4\pi {\epsilon }_0\frac{R}{\mathrm{\infty }}=0$$

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