Question

A very long insulating cylinder of charge of radius $\mathbf{2}\mathbf{.}\mathbf{50}$cm carries a uniform linear density of $\mathbf{15}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{\text{nC/m}}$. If you put one probe of a voltmeter at the surface, how far from the surface must the other probe be placed so that the voltmeter reads $\mathbf{175}$V.

Short answer

The second probe should be placed at a distance of 2.3 cm from the surface of the cylinder to make the voltmeter measure 175 V.

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 chargedensity distributed across the wire.

Determine the distance between two probes of the voltmeter

The potential difference between two points due to a charged cylinder is given by:

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

Where $r_a$and $r_b$are the distances between the cylinder's axis and the point where the potential is measured.

Here, $r_a=R=0.025\text{m}$and the term $\frac{1}{2\pi {\epsilon }_0}=18.0\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$

The target now is to get $r_b$where the distance $r_a$, in the problem equals the radius of the cylinder $r_a=R=0.025\text{m}$. And the term $\frac{1}{2\pi {\epsilon }_0}=18.0\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}$. But, the distance $r_b$is the distance between the axis of the cylinder and the point b, so the distance between the two probes of the voltmeter will be,

$d=r_b-R$

Determine the distance where the second probe should be placed

Plug the values into the equation, and we get,

$$\begin{gathered} \Delta V=\frac{\lambda }{2\pi {\epsilon }_0}\ln \frac{r_b}{r_a} \\ 175\text{V}=\left(18.0\times {10}^9\text{N}·{\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(15.0\times {10}^{-9}\text{C/m}\right)\ln \left(\frac{r_b}{0.025\text{m}}\right) \\ \ln \left(\frac{r_b}{0.025m}\right)=0.648 \\ {r}_b=0.025\text{m}\times e^{0.648} \\ =0.048\text{m} \end{gathered}$$

Now, plug this value to get the distance as:

$$\begin{gathered} d=r_b-R \\ =4.8\text{cm}-2.5\text{cm} \\ =2.3\text{cm} \end{gathered}$$

Thus, the second probe should be placed at a distance of 2.3 cm from the surface of the cylinder to make the voltmeter measure 175 V.