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

Correct Answer: 16

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.