Question

A dipole consists of two charges $+\mathbf{6}\mathbf{nC}$and $-\mathbf{6}\mathbf{nC}$, held apart by a rod of length $\mathbf{3}\mathbf{mm}$, as shown in Figure 13.71. (a) What is the magnitude of the electric field due to the dipole at location A, $\mathbf{5}\mathbf{cm}$from the center of the dipole? (b) What is the magnitude of the electric field due to the dipole at location B, $\mathbf{5}\mathbf{cm}$from the center of the dipole?

Short answer

(a) The magnitude of the electric field at location A is .$2592\text{\hspace{0.33em}N/C}$

(b) Themagnitude of the electric field at location A is .$1296\text{\hspace{0.33em}N/C}$

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The charge of the dipole is,$q=\pm 6\text{\hspace{0.33em}nC}\times \left(\frac{\text{1\hspace{0.33em}C}}{{\text{10}}^9\text{\hspace{0.33em}nC}}\right)=\text{6}\times {\text{10}}^{-\text{9}}\text{\hspace{0.33em}C}$
• The distance between the dipole is,$s=3\text{\hspace{0.33em}mm}\times \left(\frac{\text{1\hspace{0.33em}m}}{\text{1000\hspace{0.33em}mm}}\right)=\text{0.003\hspace{0.33em}m}$
• The distance between the dipole’s center to the location A is,$r_1=5\text{\hspace{0.33em}cm}\times \left(\frac{\text{1\hspace{0.33em}m}}{\text{100\hspace{0.33em}cm}}\right)=\text{0.05\hspace{0.33em}m}$
• The distance between the dipole’s center to the location B is,$r_2=5\text{\hspace{0.33em}cm}\times \left(\frac{\text{1\hspace{0.33em}m}}{\text{100\hspace{0.33em}cm}}\right)=\text{0.05\hspace{0.33em}m}$

Significance of the electric field due to a dipole

The electric field is beneficial for a charged particle to exert force on another charged particle. The magnitude of the electric field due to dipole is directly proportional to the charge and the distance of separation of the dipoles and inversely proportional to the distance between the center of the dipole to the point of the electric field.

(a) Determination of the magnitude of the electric field at location A

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

$E_1=k\frac{2qs}{{r_1}^3}$

Here, $k$is the electric field constant,$q$ is the charge of the dipole,$s$ is the distance between the dipole and is the distance between the dipole’s center to the location A.

Substitute the values in the above equation.

$\begin{array}{c}{E}_1=\left(9\times {10}^9\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{2\left({\text{6×10}}^{\text{-9}}\text{\hspace{0.33em}C}\right)\left(\text{0.003\hspace{0.33em}m}\right)}{{\left(0.05\text{\hspace{0.33em}m}\right)}^3}\\ =\left(9\times {10}^9\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left({\text{3.6×10}}^{\text{-11}}\text{\hspace{0.33em}C}\cdot \text{m}\right)}{\left({\text{1.25×10}}^{\text{-4}}{\text{\hspace{0.33em}m}}^3\right)}\\ =\left(9\times {10}^9\text{\hspace{0.33em}N}\cdot {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.88\times {10}^{-7}{\text{\hspace{0.33em}C/m}}^{\text{2}}\right)\\ =2592\text{\hspace{0.33em}N/C}\end{array}$

Thus, the magnitude of the electric field at location A is .$2592\text{\hspace{0.33em}N/C}$

(b) Determination of the magnitude of the electric field at location B

As the location B is at the perpendicular direction of the dipole, then the magnitude of the electric field at the location B will be half of the magnitude of the electric field at the location A. The reason is that due to staying in a perpendicular direction, the electric field of the point and the dipole cancels each other. Hence, the magnitude of the electric field at location B is .$2592\text{\hspace{0.33em}N/C/2}=\text{1296\hspace{0.33em}N/C}$

Thus, the magnitude of the electric field at location A is .$1296\text{\hspace{0.33em}N/C}$