Question

An electric field is applied to a solution containing bromide ions. As a result, the ions move through the solution with an average drift speed of $\mathbf{3}\mathbf{.}\mathbf{7}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}\mathbf{m}\mathbf{/}\mathbf{s}$. The mobility of bromide ions in solution is $\mathbf{8}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\left(m/s\right)\left(N/C\right)$. What is the magnitude of the net electric field inside the solution?

Short answer

The magnitude of the net electric field inside the solution is $4.57\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Identification of the given data

The given data is listed below as,

• The average drift speed of the bromide ions is,$\mathrm{v}=3.7\times {10}^{-7}\mathrm{m}/\mathrm{s}$.
• The mobility of the bromide ion is,$\mathrm{\mu }=8.7\times {10}^{-8}\left(\mathrm{m}/\mathrm{s}\right)/\left(\mathrm{N}/\mathrm{C}\right)$.

Significance of the drift velocity

Drift velocity can be obtained by taking the product of the electric field and the mobility andthe equation of the drift velocity gives the magnitude of the net electric field.

Determination of the magnitude of the net electric field

The expression for the drift velocity is as follows,

$V=\mu E$

Here,$\mu$is the mobility of the bromide ion and E is the electric field.

Substitute all the values in the above expression.

$$\begin{gathered} 3.7\times {10}^{-7}\mathrm{m}/\mathrm{s}=5.2\times {10}^{-8}\left(\mathrm{m}/\mathrm{s}\right)\left(\mathrm{N}/\mathrm{C}\right)\times \mathrm{E} \\ \mathrm{E}=\frac{3.7\times {10}^{-7}\mathrm{m}/\mathrm{s}}{5.2\times {10}^{-8}\left(\mathrm{m}/\mathrm{s}\right)\left(\mathrm{N}/\mathrm{C}\right)} \\ =4.57\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the magnitude of the net electric field inside the solution is 4.5N/C .