Q3Q - Fundamentals Of Physics

Question

Figure 26-17 shows a rectangular solid conductor of edge lengths L, 2L, and 3L. A potential difference V is to be applied uniformly between pairs of opposite faces of the conductor as in Fig. 26-8b. (The potential difference is applied between the entire face on one side and the entire face on the other side.) First V is applied between the left–right faces, then between the top–bottom faces, and then between the front–back faces. Rank those pairs, greatest first, according to the following (within the conductor): (a) the magnitude of the electric field, (b) the current density, (c) the current, and (d) the drift speed of the electrons.

Step-by-Step Solution

Answer

a) Rank of the pairs according to the magnitude of electric field is $E_{TB}>E_{FB}>E_{LR}$.

b) Rank of the current density is $J_{TB}>J_{FB}>J_{LR}$.

c) Rank of the current of the electrons is $I_{TB}>I_{FB}>I_{LR}$

d) Rank of the drift speed of the electrons is $v_{dTB}>v_{dFB}>v_{dLR}$.

Step 1: The given data

Figure 26-17 is the rectangular solid conductor of edge lengths L, 2L or 3L.

Step 2: Understanding the concept of the electric field, current density and drift velocity

To explain the electrostatic force between the two charges, we assume that the charges create an electric field around them. The magnitude of electric field E set up by the electric charge q at a distance r is given as,

$\mathit{E}\mathbf{=}\frac{\mathbf{q}}{\mathbf{4}\mathbf{\tau }\mathbf{\tau }{\mathbf{\epsilon }}_{\mathbf{0}}{\mathbf{r}}^{\mathbf{2}}}$

The current density is electric current per unit cross-section area at a given point. The drift velocity of the electrons is the average velocity attained by the electrons.

 

We can use the formulae for the electric field, current density, current, and drift velocity to rank the given pairs of faces according to an electric field, current density, current, and the drift speed of the electrons.

 

Formulae:

The strength of the electric field, $E=\frac{V}{d}$                                                                        …(i)

Here,E is the electric field,V is the potential difference, and is the separation between the two plates.

The current density of the material has a uniform current,$\mathrm{J}=\mathrm{IA}$                               …(ii)

Here,J is the current density,I is current, and A is the area of the cross-section.

The drift velocity of electrons,$V_d=\frac{J}{ne}$                                                                       …(iii)

Here,$v_d$ is the drift velocity of the electrons,J is the current density,n is the number of electrons, and is the charge of electrons.

Step 3: (a) Calculation of the rank due to the magnitude of the electric field

From equation (i), the electric field depends on the separation between the two faces of the conductor.

For left-right faces, the electric field through this face is given by,

$E_{LR}=\frac{V}{3L}$

For top –bottom faces, the electric field through this face is given by,

$E_{TB}=\frac{V}{L}$

For front-back faces, the electric field through this face is given by,

$E_{FB}=\frac{V}{2L}$

From this, we can rank the pairs according to the magnitude of electric fields as,

$E_{TB}>E_{FB}>E_{LR}$

Step 4: (b) Calculation of the rank according to current density:

Current density depends on area here, because the charge flow will be same, considering equation (ii).

For left-right faces, the current density from this face is given by:

$$\begin{gathered} J_{LR}=I\left(2I\times L\right) \\ =2I{L}^2 \end{gathered}$$

For top –bottom faces, the current density from this face is given by:

$$\begin{gathered} J_{TB}=I\left(2L\times 3L\right) \\ =6I{L}^2 \end{gathered}$$

For front-back faces, the current density from this face is given by:

$$\begin{gathered} J_{FB}=I\left(L\times 3L\right) \\ =3I{L}^2 \end{gathered}$$

From the above equations, we can rank the pairs according to the current density as,

$J_{TB}>J_{FB}>J_{LR}$

Step 5: (c) Calculation of the rank according to the current flowing through it

For left-right faces, the current through this face using equation (ii) and above values is given by:

$I_{LR}=J\left(2L^2\right)$

For top –bottom faces, the current through this face using equation (ii) and above values is given by:

$I_{TB}=J\left(6L^2\right)$

For front-back faces, the current through this face using equation (ii) and above values is given by:

$I_{FB}=J\left(3L^2\right)$

From the above equations, we can rank the pairs according to the current as,

$I_{TB}>I_{FB}>I_{LR}$

Step 6: (d) Calculation of the rank according to the drift speed of the electrons

Here, the drift velocity is directly proportional to the current density considering equation (iii).

From part b), we can write,$J_{TB}>J_{FB}>J_{LR}$ 

Hence, the rank value according to their speeds is $v_{dTB}>v_{dFB}>v_{dLR}$.