Question

Two dipoles are oriented as shown in Figure 13.72. Each dipole consists of two charges +qand -q, held apart by a rod of length s, and the center of each dipole is a distance dfrom location A. If$\mathbf{=}\mathbf{2}\mathbf{nC}$, $\mathbf{s}\mathbf{=}\mathbf{1}\mathbf{mm}$and d=8cm, what is the electric field at location A? (Hint: Draw a diagram and show the direction of each dipole’s contribution to the electric field on the diagram.)

Short answer

The electric field at the location A is

Step-by-step solution

Identification of the given data

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{data}\mathrm{can}\mathrm{be}\mathrm{listed}\mathrm{below}\mathrm{as}: \\ •\mathrm{The}\mathrm{charges}\mathrm{on}\mathrm{the}\mathrm{dipoles}\mathrm{are},+\mathrm{q}=2\mathrm{nC}\mathrm{and}-\mathrm{q}=2\mathrm{nC} \\ •\mathrm{The}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{rod}\mathrm{is},\mathrm{s}=1\mathrm{mm}\times \left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=1\times {10}^{-3}\mathrm{m} \\ •\mathrm{The}\mathrm{distance}\mathrm{of}\mathrm{the}\mathrm{dipoles}\mathrm{from}\mathrm{location}\mathrm{A}\mathrm{is},\mathrm{d}=8\mathrm{cm}\times \left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=8\times {10}^{-2}\mathrm{m} \end{gathered}$$

Significance of the magnitude of the electric field

The magnitude of the electric field is directly proportional to the dipole moment and inversely proportional to the distance and the length of an object.

Determination of the electric field due to one dipole

The diagram of the direction of the dipoles has been provided below:

The equation of the magnitude of the electric field of the parallel dipole is expressed as:

$E_{║}=k\frac{2p}{d\sqrt{d^2+s^2}}$ …(i)

Here, k is the electric field constant that has the value of $9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$, p is the dipole moment, d is the distance of the dipoles from the location A and is the length of the rod.

The distance of the dipoles from the location A is much bigger than the length of the rod. Hence,$\mathrm{d}\gg \mathrm{s}$.

Then the above equation can be reduced as:

$E_{║}=k\frac{2p}{d^3}$

The equation of the dipole moment is expressed as:

p=qs

Here, q is the charge of the dipoles and s is the length of the rod.

Substitute the value in the equation (i).

$$\begin{gathered} {\mathrm{E}}_{║}=\mathrm{k}\frac{2\mathrm{qs}}{{\mathrm{d}}^3} \\ \mathrm{Substitute}\mathrm{all}\mathrm{the}\mathrm{values}\mathrm{in}\mathrm{the}\mathrm{above}\mathrm{equation}. \\ {\mathrm{E}}_{║}=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{2\times 2\times {10}^{-9}\mathrm{C}\times 1\times {10}^{-3}\mathrm{m}}{{\left(8\times {10}^{-2}\mathrm{m}\right)}^3} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{4\times {10}^{-12}\mathrm{C}.\mathrm{m}}{5.12\times {10}^{-4}{\mathrm{m}}^3} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times 7.81\times {10}^{-59}\mathrm{C}/{\mathrm{m}}^2 \\ =70.3\mathrm{N}/\mathrm{C} \end{gathered}$$

Determination of the electric field due to another dipole

The equation of the magnitude of the electric field of the perpendicular dipole is expressed as:

$E_{┴}=k\frac{p}{d\sqrt{d^2+s^2}}$

…(ii)

Here, k is the electric field constant that has the value $9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2$, p is the dipole moment, d is the distance of the dipoles from the location A and s is the length of the rod.

The distance of the dipoles from the location A is much bigger than the length of the rod. Hence,$d\gg s$.

Then the above equation can be reduced as:

$E_{┴}=k\frac{p}{d^3}$

The equation of the dipole moment is expressed as:

p=qs

Here, q is the charge of the dipoles and s is the length of the rod.

Substitute the value in the equation (ii).

$$\begin{gathered} {\mathrm{E}}_{┴}=9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{2\times 2\times {10}^{-9}\mathrm{C}\times 1\times {10}^{-3}\mathrm{m}}{{\left(8\times {10}^{-2}\mathrm{m}\right)}^3} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{2\times {10}^{-12}\mathrm{C}.\mathrm{m}}{5.12\times {10}^{-4}{\mathrm{m}}^3} \\ =9\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times 3.906\times {10}^{-9}\mathrm{C}/{\mathrm{m}}^2 \\ =35.1\mathrm{N}/\mathrm{C} \end{gathered}$$

Determination of the electric field at location Ac

As the distance of the dipoles from the location A is much higher than the length of the rod, then the dipole in the x axis is linear. Hence, the dipole on the x axis is 0. Moreover, the dipole on the z axis is also 0 as it is on the xy plane.

The equation of the magnitude of the electric field at the location A is expressed as:

$E=E_{║}+E_{┴}$

Here, $E_{║}$is the parallel electric field and$E_{┴}$ is the perpendicular electric field.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{E}=70.3\mathrm{N}/\mathrm{C}+35.1\mathrm{N}/\mathrm{C} \\ =105.4\mathrm{N}/\mathrm{C} \end{gathered}$$

Thus, the electric field at the location A is.