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# The Electric Dipole and Bipole Fields Along the Dipole Axis

EANDM-KVY5GR

Figure 1 below depicts an electric dipole, in which the separation between the positive and negative charges is equal to $\alpha = 3.9 \text{ pm}$, and the magnitude of each charge is equal to $q = 10e$, where $e = 1.602 \times 10^{-19} \text{ C}$ is the electronic charge.

Figure 2 depicts an electric bipole, which is identical to the electric dipole, with the same charge magnitudes and charge separation. The only difference between the dipole and the bipole is that, in the bipole, the negative charge has been replaced with a positive charge. The field point $P$ in both figures is along the dipole/bipole axis, which is collinear with the $\hat{y}$ axis.

Using the notation of figure 3, the electric field of a dipole at point $P$ is equal to:

$$\vec{E}_{dip} = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1^2} - \frac{1}{r_2^2}\right) \hat{y}$$

The electric field of a bipole at point $P$ is equal to:

$$\vec{E}_{dip} = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r_1^2} + \frac{1}{r_2^2}\right) \hat{y}$$

What is the ratio of the magnitude of the electric dipole field to the magnitude of the electric bipole field:

$$R = \frac{E_{dip}}{E_{bip}}$$

...at a distance of $z = 4.3 \text{ nm}$? ​

A

$R = 1$

B

$R = 9.07 \times 10^{-4}$

C

$R = 1.81 \times 10^{-3}$

D

$R = 1100$