91Ó°ÊÓ

The given data is listed below as:

• The value of the charge is,.$Q=1\text{ n°ä}\left({\text{1×10}}^{\text{-9}}\text{â€ÁÝ}\right)$
• The dipoles are separated by a distance of .$\text{s}=\text{0.3 m³¾}=\text{0.3 m³¾}×\frac{\text{1 m}}{\text{1000 m³¾}}={\text{0.3×10}}^{−\text{3}}\text{ m}$
• The distance between the charge and the location A is, .$R=\text{20 c³¾}=\text{20 c³¾}×\frac{\text{1 m}}{\text{100 c³¾}}={\text{20×10}}^{−\text{2}}\text{ m}$
• The distance between the dipoles and the location A is,.$r=1\text{0 c³¾}=1\text{0 c³¾}×\frac{\text{1 m}}{\text{100 c³¾}}=1{\text{0×10}}^{−\text{2}}\text{ m}$

The electric field of a charged particle is directly proportional to the charge and inversely proportional to the square of the distance of the charge to the location. Moreover, the electric field due to dipole is directly proportional to the charge of the dipole and inversely proportional to the cube of the distance from the dipole to the location.

As the dipole is located at below the point A and as the net electric field at the location A is zero, it states that the right end of the dipole is positively charged. The reason the right end is positively charged is that due to the positively charged particle, the net electric field is zero, otherwise it would have a value.

Thus, the right end of the dipole is positively charged.

The equation of the magnitude of the electric field due to the charged particle is expressed as:

$E_1=k\frac{Q}{R^2}$

Here, $k$is the electric field constant,$Q$is the value of the charge and $R$is the distance between the charge and the location A.

The equation of the magnitude of the electric field due to the dipoles is expressed as:

$E_2=k\frac{−qs}{r^3}$

Here, $k$is the electric field constant, $q$is the value of the charge and $r$is the distance between the dipole and the location A.

According to the question, the net electric field at the location A is zero. Hence, the equation of the net electric field is expressed as:

$E=E_1+E_2$

Here, $E_1$is the magnitude of the electric field due to the charged particle and$E_2$is the magnitude of the electric field due to the dipole.

The above equation can also be expressed as:

$\begin{array}{c}0=k\frac{Q}{R^2}−k\frac{qs}{r^3}\\ k\frac{qs}{r^3}=k\frac{Q}{R^2}\\ \frac{qs}{r^3}=\frac{Q}{R^2}\end{array}$

Substitute the values in the above equation.

$\begin{array}{c}\frac{q\left(\text{0.3}×{\text{10}}^{\text{-3}}\text{ m}\right)}{{\left(\text{10}×{\text{10}}^{\text{-2}}\text{ m}\right)}^3}=\frac{\left(\text{1}×{\text{10}}^{\text{-9}}\text{â€ÁÝ}\right)}{{\left(\text{20}×{\text{10}}^{\text{-2}}\text{ m}\right)}^2}\\ \frac{q\left(\text{0.3}×{\text{10}}^{\text{-3}}\text{ m}\right)}{\text{1}×{\text{10}}^{\text{-3}}{\text{ m}}^3}=\frac{\left(\text{1}×{\text{10}}^{\text{-9}}\text{â€ÁÝ}\right)}{{\text{0.04 m}}^{\text{2}}}\\ q\left({\text{0.3 m}}^2\right)=\left(\text{2.5}×{\text{10}}^{\text{-8}}{\text{â€ÁÝ/m}}^2\right)\\ q=8.3×{10}^{−8}\text{â€ÁÝ}\end{array}$

Thus, the value of the charge$\text{q}$ is.$8.3×{10}^{−8}\text{â€ÁÝ}$