Question

A very long wire carries a uniform linear charge density $\mathit{\lambda }$. Using a voltmeter to measure potential difference, you find that when one probe of the meter is placed $\mathbf{2}\mathbf{.}\mathbf{50}$cm from the wire and the other probe is $\mathbf{1}\mathbf{.}\mathbf{00}$farther from the wire, the meter reads $\mathbf{575}$V.

(a) What is $\mathit{\lambda }$?

(b) If you now place one probe at$\mathbf{3}\mathbf{.}\mathbf{50}$ cm from the wire and the other probe $\mathbf{1}\mathbf{.}\mathbf{00}$cm farther away, will the voltmeter read$\mathbf{575}$ V? If not, will it read more or less than$\mathbf{575}$ V? Why?

(c) If you place both probes $\mathbf{3}\mathbf{.}\mathbf{50}$cm from the wire but $\mathbf{17}\mathbf{.}\mathbf{0}$ cm from each other, what will the voltmeter read?

Short answer

1. The value $\mathit{\lambda }$of is$\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{\text{C}}$
2. The voltmeter will read 429 V because the potential energy between them is less than the previous.
3. The voltmeter will read zero volts.

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.

Calculation of λ

(a)

Rearranging the potential law for infinite wire as:

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

Given data:

$V_a-V_b=575\text{V}$

$r_b=2.5+1=3.5\text{cm}$

$r_a=2.5\text{cm}$

On substitution in the above expression, we get,

$$\begin{gathered} \lambda =\frac{575\times 2\pi \times 8.85\times {10}^{-12}}{\ln \left(3.5/2.5\right)} \\ =9.5\times {10}^{-8}\text{C} \end{gathered}$$

Thus, the value of $\mathit{\lambda }$is $\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{\text{C}}$

Determine the reading of the voltmeter if the probes are placed at a particular distance from the wire

(b)

Here, $\lambda =9.5\times {10}^{-8}\text{C}$

So, on substitution,

$$\begin{gathered} V_a-V_b=\frac{9.5\times {10}^{-8}}{2\pi \times 8.85\times {10}^{-12}}\ln \left(\frac{4.5}{3.5}\right) \\ =429\text{V} \end{gathered}$$

Thus, the voltmeter reads less than $575\text{V}$.

The potential difference is decreased because we need less potential to bring the test charge from infinity to each point, so the potential difference is decreased between them.

Determine the reading of the voltmeter if the probes are placed a particular distance away from each other and the same distance from the wire.

(c)

Since both probes are at the same distance from the wire as, $r_b=r_a$then the potential difference must be equal to zero because the vertical distance doesn’t matter.

Thus, The value $\lambda$is $\mathbf{9}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathbf{\text{C}}$that the voltmeter will read because the potential energy between them is less than the previous, and the voltmeter will read zero volt if the probes are placed at an equal distance from the wire.