Question

A very long insulating cylindrical shell of radius $\mathbf{6}\mathbf{.}\mathbf{00}\mathbf{}$cm carries a charge of linear density $\mathbf{8}\mathbf{.}\mathbf{50}\mathit{\mu }\mathit{C}\mathbf{/}\mathit{m}$spread uniformly over its outer surface. What would a voltmeter read if it were connected between

(a) the surface of the cylinder and a point $\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}$above the surface, and

(b) the surface and the point $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{\text{cm}}$ from the central axis of the cylinder?

Short answer

1. The voltmeter reading is $78.0\text{kV}$.
2. The voltmeter reading is zero.

Step-by-step solution

Potential law for infinite wire

Let a and b be two points, and the potential at those points is V_a and V_b, respectively. The potential difference between two points for an infinite wire can be written as:

$V_a-V_b=\frac{\lambda }{2\pi {\epsilon }_0}\ln \frac{r_b}{r_a}$

Here r_a and r_b are the distance of the wire from points a and b, respectively. $\mathit{\lambda }$is the linear charge density distributed across the wire.

Determine the voltmeter reading in part (a)

Given data:

$\lambda =8.5\times {10}^6\text{C/m}$

$R=6.0\text{cm}$

$r=10.0\text{cm}$

From the potential law for infinite wire, substitute the given values in the equation,

$$\begin{gathered} V=\frac{8.5\times {10}^{-6}}{2\pi \times 8.85\times {10}^{-12}}\ln \left(\frac{10}{6}\right) \\ =78.0\text{kV} \end{gathered}$$

Thus, the voltmeter reading is $78.0\text{kV}$.

Determine the voltmeter reading in part (b)

The electric field inside the cylinder is equal to zero. So, depending on that, the voltage inside the cylinder is constant and equal to the potential at the surface, so it’s equal to zero because $r=R$and $\ln 1=0$.

Thus, the voltmeter reading is zero.