Question

A point dipole is located at the origin in some orientation. The electric field at the point $\left( {10cm,10cm} \right)$ on the x-y plane is measured to have a magnitude $1.0 \times {10^{ - 3}}V/m$ . What will be the magnitude of the electric field at the point $\left( {20cm,20cm} \right)$ ?
(A) $5.0 \times {10^{ - 4}}V/m$
(B) $2.5 \times {10^{ - 4}}V/m$
(C) It will depend on the orientation at the dipole
(D) $1.25 \times {10^{ - 4}}V/m$

Answer 1

Electric dipoles will be having two charges. Just like magnetic dipoles, we can consider electric dipoles. In a magnetic dipole, the magnetic moment vector goes from the South Pole to the North Pole inside the magnet. We resolve that along axial and equatorial to get the magnetic fields. Similarly in an electric dipole, the dipole moment vector goes from negative charge to positive charge.

Complete Step-by-step solution:
Given, the electric field intensity at the point $\left( {10cm,10cm} \right)$ say $E$ is $1.0 \times {10^{ - 3}}V/m$ .
First of all, we have to find the distance of the dipole from the origin. Let the distance of the dipole from the origin is $r$ .
Since we know the formula for finding the distance of origin from the given point is;
$r = \sqrt {{{\left( {x - 0} \right)}^2} + {{\left( {y - 0} \right)}^2}} = \sqrt {{x^2} + {y^2}}$ ………. $\left( 1 \right)$
Put the values in the above equation,
$\Rightarrow r = \sqrt {{{10}^2} + {{10}^2}}$
$\Rightarrow r = \sqrt {100 + 100} = \sqrt {200}$
On further solving we get,
$\Rightarrow r = 10\sqrt 2 cm$
To convert the distance into meters we have to divide it with $\;100$ .
$\Rightarrow r = 0.1\sqrt 2 m$
The electric field due to a dipole at any point in space is given by,
$E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{\rho }{{{r^3}}}$ ……… $\left( 2 \right)$
Where,
$\rho$ is the dipole moment of the dipole
$r$ is the distance from the point of the dipole
On solving the equation for getting the value of dipole moment
$\Rightarrow \rho = 4\pi {\varepsilon _0}{r^3}\left( E \right)$ ……….. $\left( 3 \right)$
Now, we have to calculate the electric field intensity at the point $\left( {20cm,20cm} \right)$ .
Let the distance of the dipole from origin now be $\;r'$
Since we know the formula to find the distance of a given point from the origin. Substitute all the values in the equation $\left( 1 \right)$ we get,
$\Rightarrow r' = \sqrt {{{20}^2} + {{20}^2}}$
$\Rightarrow r' = \sqrt {400 + 400} = \sqrt {800}$
On further solving we get,
$\Rightarrow r' = 20\sqrt 2 cm = 0.2\sqrt 2 m$
Since the dipole moment of the dipole remains the same at any point.
Therefore, the electric field intensity of the dipole at the point $\left( {20,20} \right)$ is,
$E' = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{\rho }{{{{r'}^3}}}$
Substitute the value of $\rho$ from the equation $\left( 3 \right)$ we can write,
$\Rightarrow E' = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{4\pi {\varepsilon _0}{r^3}\left( E \right)}}{{{{r'}^3}}}$
$\Rightarrow E' = \dfrac{{{r^3}E}}{{{{r'}^3}}}$ ……….. $\left( 4 \right)$
Substitute all the values to get the final answer,
$\Rightarrow E' = \dfrac{{{{\left( {0.1\sqrt 2 } \right)}^3} \times 1 \times {{10}^{ - 3}}}}{{{{\left( {0.2\sqrt 2 } \right)}^3}}}$
$\Rightarrow E' = \dfrac{1}{8} \times {10^{ - 3}} = 1.25 \times {10^{ - 4}}V/m$
Therefore, the magnitude of the electric field at the point $\left( {20cm,20cm} \right)$ is $1.25 \times {10^{ - 4}}V/m$ .
Hence, the correct answer is option (D).

Note:
Electric field is defined as the electric force per unit charge. The direction of the field taken to be the direction of the force it would exert on a positive test charge. The electric field radially outward from a positive charge and radially toward a negative point charge.