91Ó°ÊÓ

The given data is listed below as:

• The charge of the dipole is,$q=Â\pm 6\text{ n°ä}×\left(\frac{\text{1â€ÁÝ}}{{\text{10}}^9\text{ n°ä}}\right)=\text{6}×{\text{10}}^{−\text{9}}\text{â€ÁÝ}$
• The distance between the dipole is,$s=3\text{ m³¾}×\left(\frac{\text{1 â¶Ä‹m}}{\text{1000 m³¾}}\right)=\text{0.003 m}$
• The distance between the dipole’s center to the location A is,$r_1=5\text{ c³¾}×\left(\frac{\text{1 â¶Ä‹m}}{\text{100 c³¾}}\right)=\text{0.05 m}$
• The distance between the dipole’s center to the location B is,$r_2=5\text{ c³¾}×\left(\frac{\text{1 â¶Ä‹m}}{\text{100 c³¾}}\right)=\text{0.05 m}$

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.

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.

$\begin{array}{c}{E}_1=\left(9×{10}^9\text{ â¶Ä‹N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{2\left({\text{6×10}}^{\text{-9}}\text{â€ÁÝ}\right)\left(\text{0.003 m}\right)}{{\left(0.05\text{ m}\right)}^3}\\ =\left(9×{10}^9\text{ â¶Ä‹N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\frac{\left({\text{3.6×10}}^{\text{-11}}\text{â€ÁÝ}â‹\dots \text{m}\right)}{\left({\text{1.25×10}}^{\text{-4}}{\text{ m}}^3\right)}\\ =\left(9×{10}^9\text{ â¶Ä‹N}â‹\dots {\text{m}}^{\text{2}}{\text{/C}}^{\text{2}}\right)\left(2.88×{10}^{−7}{\text{â€ÁÝ/m}}^{\text{2}}\right)\\ =2592\text{ N/°ä}\end{array}$

Thus, the magnitude of the electric field at location A is .$2592\text{ N/°ä}$

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 .$2592\text{ N/°ä/2}=\text{1296 N/°ä}$

Thus, the magnitude of the electric field at location A is .$1296\text{ N/°ä}$