You place a neutral block of nickel near a small glass sphere that has a charge of $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{C}$uniformly distributed over its surface, as shown in Figure 14.92.

(a) About how long do you have to wait to make sure that the mobile electron sea inside the nickel block has reached equilibrium? (1) Less than a nanosecond $\mathbf{(}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathbf{s}\mathbf{)}$, (2) Several hours, (3) About $\mathbf{1}\mathit{s}$, (4) About $\mathbf{10}\mathbf{}\mathit{m}\mathit{i}\mathit{n}$(b) In equilibrium, what is the average drift speed of the mobile electrons inside the nickel block? (1) About $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$, (2) About $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathit{m}\mathbf{/}\mathit{s}$, (3) $\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$(c) In the equation $\overline{\mathbf{v}}\mathbf{=}\mathit{u}\mathit{E}$, what is the meaning of the symbol $\mathit{u}$? (1) The density of mobile electrons inside the metal, in localid="1657175774793" $\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}\mathbf{/}{\mathit{m}}^{\mathbf{3}}$, (2) The mobility of an electron inside the metal, in $\mathit{m}\mathbf{/}\mathit{s}\mathbf{/}\mathit{N}\mathbf{/}\mathit{C}$, (3) The time it takes a block of metal to reach equilibrium, in seconds

The given data is listed below as:

The charge of the neutral block is $\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{8}}\mathbf{}\mathit{C}$.

Drift mainly refers to the slow movement of an object towards another object. The drift speed is described as the average velocity of the electrons which is helpful for the electrons to drift inside an electric field.

Option 2 states that it will take several hours to make sure that the mobile electron sea inside the nickel block has reached equilibrium. This option is incorrect as after polarization, the metal reaches in equilibrium in less than a nanosecond.

Option 3 states that it will take $\mathbf{1}\mathbf{}\mathit{s}$to make sure that the mobile electron sea inside the nickel block has reached equilibrium. This option is incorrect as after polarization, the metal reaches in equilibrium in less than a nanosecond.

Option 4 states that it will take $\mathbf{10}\mathbf{}\mathit{m}\mathit{i}\mathit{n}$to make sure that the mobile electron sea inside the nickel block has reached equilibrium. This option is incorrect as after polarization, the metal reaches in equilibrium in less than a nanosecond.

Option 1 states that it will take less than a nanosecond $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{s}$to make sure that the mobile electron sea inside the nickel block has reached equilibrium. This option is correct as after polarization, the metal reaches in equilibrium in less than a nanosecond. The reason for taking the less time is that the charges is mainly needed to be displaced in a small distance.

Thus, the individual has to wait less than a nanosecond to make sure that the mobile electron sea inside the nickel block has reached equilibrium and option 1 is correct.

Option 1 states that about $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{s}$is the average drift speed of the mobile electrons inside the metal block, in equilibrium. This option is incorrect as equilibrium means the drift velocity will be 0 in the absence of the mobile charges.

Option 2 states that about $\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{9}}\mathbf{}\mathit{s}$ is the average drift speed of the mobile electrons inside the metal block, in equilibrium. This option is incorrect as equilibrium means the drift velocity will be 0 in the absence of the mobile charges.

Option 3 states that about $\mathbf{0}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$ is the average drift speed of the mobile electrons inside the metal block, in equilibrium. This option is correct as equilibrium means the drift velocity will be 0 in the absence of the mobile charges.

Thus, in equilibrium, the average drift speed of the mobile electrons inside the nickel block is 0 and option 3 is correct.

The equation given in the question is expressed as:

$\stackrel{\boxed{\mathbf{∆}}}{\mathbf{v}}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{E}$

Here, $\stackrel{\boxed{\mathbf{∆}}}{\mathbf{v}}\mathbf{}$is the velocity that is expressed as the unit of $\mathit{m}\mathbf{/}\mathit{s}$and $\mathit{E}$is described as the electric field that is expressed as $\mathit{N}\mathbf{/}\mathit{C}$.

With the help of the units, the unit of $\mathit{u}$can be calculated as:

$$\begin{gathered} \mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{=}\mathit{u}\mathbf{(}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{u}\mathbf{=}\mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{/}\mathbf{(}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)} \end{gathered}$$

Statement 1 states that the density of the electrons inside the metal is $\mathit{e}\mathit{l}\mathit{e}\mathit{c}\mathit{t}\mathit{r}\mathit{o}\mathit{n}\mathit{s}\mathbf{/}{\mathit{m}}^{\mathbf{-}\mathbf{3}}$. This statement is incorrect as from the above equation, the density cannot be identified.

Statement 3 states that the time it takes a block of metal to reach equilibrium, in seconds. This statement is incorrect as from the above equation, the time taken cannot be identified.

Statement 2 states that the mobility of an electron inside the metal, in $\mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{/}\mathbf{(}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)}$. With the help of the equation gathered above, it has been observed that the statement 2 correctly explains the mobility of the electrons inside the metal.

Thus, the meaning of the symbol $\mathit{u}$is that the mobility of an electron inside the metal, in $\mathbf{(}\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}\mathbf{/}\mathbf{(}\mathbf{N}\mathbf{/}\mathbf{C}\mathbf{)}$and option 2 is correct.