Question

The electric field at point p due to an electric dipole is E. The electric field at point R on equitorial line will be$\frac{E}{x}$. The value of x :

Answer 1

Given:
- Electric field at point $P$ on the axial line: $E_P = E = \frac{2Kp}{r^3}$
- Electric field at point $R$ on the equatorial line: $E_R = \frac{Kp}{(2r)^3}$, where:
- $K$ is the Coulomb constant,
- $p$ is the dipole moment,
- $r$ is the distance from the dipole center to the point of observation.

Step 1: Calculate the Electric Field at $R$
The electric field at point $R$ on the equatorial line is given by:

$E_R = \frac{Kp}{(2r)^3}.$

Simplify $(2r)^3$:

$E_R = \frac{Kp}{8r^3}.$

Step 2: Compare the Electric Fields
The electric field at $P$ on the axial line is:

$E_P = \frac{2Kp}{r^3}.$

The electric field at $R$ is related to $E_P$ as:

$E_R = \frac{E_P}{x}.$

Substitute $E_P = \frac{2Kp}{r^3}$ and $E_R = \frac{Kp}{8r^3}$:

$\frac{Kp}{8r^3} = \frac{2Kp}{xr^3}.$

Step 3: Solve for $x$
Simplify the equation:

$\frac{Kp}{8r^3} = \frac{2Kp}{xr^3}.$

Cancel $Kp$ and $r^3$ (as they are non-zero):

$\frac{1}{8} = \frac{2}{x}.$

Rearrange to solve for $x$:

$x = 2 \times 8 = 16.$

Thus, the value of $x$ is 16.