Ncert Exercises & Solutions

A cylindrical capacitor has two co-axial cylinders of length \(L=15~\text{cm}\) and radii \(r_a=1.4~\text{cm}\)(inner) and \(r_b=1.5~\text{cm}\)(outer). The outer cylinder is earthed, and the inner cylinder is given a charge of \(q=3.5~\mu\text C.\) Determine the capacitance of the system and the potential of the inner cylinder. Neglect end effects. (i.e., bending of field lines at the ends)
1. \(C= 1.21 \times 10^{-10}~\text F,~V_a=2.89\times10^4~\text V\)
2. \(C= 1.38 \times 10^{-10}~\text F,~V_a=2.54 \times10^4~\text V\)
3. \(C= 1.21 \times 10^{-9}~\text F,~V_a=2.89\times10^3~\text V\)
4. \(C= 1.38 \times 10^{-9}~\text F,~V_a=2.54 \times10^3~\text V\)

Length of a co-axial cylinder I = 15 cm = 0.15 m

Radius of outer cylinder, r₁ = 1.5 cm = 0.015 m

Radius of inner cylinder, r₂ = 1.4 cm = 0.014 m

Charge on the inner cylinder, q = 3.5

μ

C = 3.5 x 10^-6 C

The capacitance of a co-axial cylinder of radii r₁ and r₂ is given by the relation

C = \frac{2 {πϵ}_{0} l}{\log_{e} \frac{r_{1}}{r_{2}}}

Where,

\begin{array}{l} ϵ_{0} = Permittivity of free space = 8 .85 \times 10^{- 12} N^{- 1} m^{- 2} C^{2} \\ ∴ C = \frac{2 π \times 8 .85 \times 10^{- 12} \times 0 .15}{2 .3026 \log_{10} (\frac{0 .15}{0 .14})} \\ = \frac{2 π \times 8 .85 \times 10^{- 12} \times 0 .15}{2 .3026 \times 0 .0299} = 1 .2 \times 10^{- 10} F \end{array}

The potential difference of the inner cylinder is given by,

\begin{aligned} V & = \frac{q}{C} \\ = \frac{3 .5 \times 10^{- 6}}{1 .2 \times 10^{- 10}} = 2 .92 \times 10^{4} V \end{aligned}

NEET Information

courses

Company

Botany Questions

Chemistry Questions

Physics Questions

Zoology Questions