Question

An electric field of magnitude $\mathbf{190}\mathbf{ }\mathbf{ }\mathbf{\text{N}}\mathbf{/}\mathbf{\text{C}}\mathbf{}$ is applied to a solution containing chloride ions. The mobility of chloride ions in solution is $\mathbf{7}\mathbf{.}\mathbf{91}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{ }\mathbf{ }\left(\text{m}/\text{s}\right)\mathbf{/}\left(\text{N}/\text{C}\right)\mathbf{}$.What is the average drift speed of the chloride ions in the solution?

Short answer

$1.5\times {10}^{-5}\text{m}/\text{s}$

Step-by-step solution

Identification of given data

• The magnitude of the electric field is, $E=190\text{N}/\text{C}$
• The mobility of chloride ions is, $\mu =7.91\times {10}^{-8} \left(\text{m}/\text{s}\right)/\left(\text{N}/\text{C}\right)$

Explanation of Drift speed

The drift speed can be determined by taking the product of the electric field and the mobility of ions. It is the average velocity attained by the charged particles in the presence of the electric field.

Determination of the average drift speed of chloride ions

The expression for the drift speed of ions is as follows,

$\nu =E\times \mu$

Here,$E$is magnitude of the electric field, and $\mu$is the mobility of chloride ions.

For $E=190\text{N}/\text{C}$and $\mu =7.91\times {10}^{-8} \left(\text{m}/\text{s}\right)/\left(\text{N}/\text{C}\right)$.

$$\begin{gathered} \nu =190N/C\left(7.91\times {10}^{-8}\left(m/s\right)/\left(N/C\right)\right) \\ =1.5\times {10}^{-5}m/s \end{gathered}$$

Thus, the average drift speed of chloride ions in the solution is $1.5\times {10}^{-5}\text{m}/\text{s}$.