Question

A hollow cylinder of radius and height has a total charge uniformly distributed over its surface. The axis of the cylindercoincides with the z axis, and the cylinder is centered atthe origin.What is the potential $V_0$ in the limit as goes to zero?Express your answer in terms of,,and $epsilon_0$ .

Answer 1

The cylinder will become a ring when the height goes to zero. Thus consider a ring with radius \(r,\) in which a charge of \(q\) is uniformly distributed. The charge is distributed along the circumference, so the linear charge density should be chosen, which will be a constant.

The linear charge density for a circle (ring) is given by,

$$ \lambda=\frac{q}{2 \pi r} $$

where \(\lambda\) is the linear charge density, \(q\) is the total charge and \(r\) is the radius.

For a small line element in the circumference of the ring, the charge is \(d q .\) Thus the charge is expressed as \(d q=\lambda d l=\lambda r d \theta,\) where \(d l\) is the small arc in which the charge \(d q\) is present.

The expression for the potential for a charge distribution is given by,

$$ V=\int \frac{d q}{4 \pi \epsilon_{0} r} $$

Substitute \(d q\) and obtain,

$$ \begin{aligned} V &=\int_{0}^{2 \pi} \frac{\lambda}{4 \pi \varepsilon_{0}} d \theta=\frac{\lambda}{4 \pi \varepsilon_{0}}[\theta]_{0}^{2 \pi} \\ &=\frac{\left(\frac{q}{2 \pi r}\right)}{4 \pi \varepsilon_{0}} \times 2 \pi \\ &=\frac{q}{4 \pi \varepsilon_{0} r} \end{aligned} $$

Answer 2

Potential due to the small ring element at a distance z fromthe origin and having width = dz is,since ln(a)-ln(b) = ln(a/b)In the limit h tendin to zero, the above becomes,