Question

What do you understand by capacity of any conductor? Find an expression for the capacity of a cylindrical condenser.

Hint:

The general strategy to find the capacitance for any geometry is a three-step process: 1. Assume a charge \(+Q\) and \(-Q\) on the conductors. 2. Calculate the electric field \(\vec{E}\) between them (usually using Gauss's law). 3. Integrate \(\vec{E}\) to find the potential difference \(V\). 4. Use the formula \(C = Q/V\).

Answer 1

Part 1: Capacity of a Conductor
Step 1: Understanding the Concept:
The capacity, or capacitance, of a conductor is its intrinsic ability to store electric charge. When charge is given to an isolated conductor, its electric potential increases.
It is observed that the potential (\(V\)) of the conductor is directly proportional to the charge (\(Q\)) given to it.
\[ Q \propto V \implies Q = CV \] Step 2: Definition:
The constant of proportionality, \(C\), is called the capacitance of the conductor. It is defined as the ratio of the charge given to the conductor to the potential raised on it.
\[ C = \frac{Q}{V} \] Thus, the capacity of a conductor is numerically equal to the amount of charge required to raise its potential by one unit. Its S.I. unit is the Farad (F).
Part 2: Expression for the Capacity of a Cylindrical Condenser
Step 1: Setup and Assumptions:
A cylindrical condenser (capacitor) consists of two coaxial hollow conducting cylinders. Let the inner cylinder have radius 'a' and the outer cylinder have radius 'b'. Let the length of the cylinders be 'L', and assume \(L \gg b>a\) so that edge effects can be neglected. The space between the cylinders is filled with a dielectric medium of permittivity \(\epsilon\). Let a charge of \(+Q\) be given to the inner cylinder and \(-Q\) to the outer cylinder.

Step 2: Calculating Electric Field:
To find the electric field \(E\) at a point at a radial distance \(r\) (\(a<r<b\)) from the axis, we consider a cylindrical Gaussian surface of radius \(r\) and length \(L\).
The linear charge density is \(\lambda = Q/L\).
According to Gauss's law:
\[ \oint \vec{E} \cdot d\vec{A} = \frac{q_{enc}}{\epsilon} \] \[ E(2\pi r L) = \frac{\lambda L}{\epsilon} = \frac{Q}{\epsilon} \] \[ E = \frac{Q}{2\pi \epsilon L r} \] The direction of the electric field is radially outwards.
Step 3: Calculating Potential Difference:
The potential difference (\(V\)) between the inner and outer cylinders is the work done in moving a unit positive charge from the outer cylinder to the inner cylinder.
\[ V = -\int_{b}^{a} \vec{E} \cdot d\vec{r} = -\int_{b}^{a} \frac{Q}{2\pi \epsilon L r} dr \] \[ V = \frac{Q}{2\pi \epsilon L} \int_{a}^{b} \frac{1}{r} dr \] \[ V = \frac{Q}{2\pi \epsilon L} [\ln(r)]_{a}^{b} \] \[ V = \frac{Q}{2\pi \epsilon L} (\ln(b) - \ln(a)) = \frac{Q}{2\pi \epsilon L} \ln\left(\frac{b}{a}\right) \] Step 4: Finding the Capacitance:
Using the definition of capacitance, \(C = Q/V\):
\[ C = \frac{Q}{\frac{Q}{2\pi \epsilon L} \ln\left(\frac{b}{a}\right)} \] \[ C = \frac{2\pi \epsilon L}{\ln(b/a)} \] If the medium is a vacuum or air, \(\epsilon = \epsilon_0\), and the expression becomes:
\[ C = \frac{2\pi \epsilon_0 L}{\ln(b/a)} \] This is the required expression for the capacity of a cylindrical condenser.