Question

What is an electric dipole? Find an expression for electric potential at any point due to an electric dipole.

Hint:

The electric potential of a dipole decreases as \( \frac{1}{r^2} \), faster than that of a single charge (\( \frac{1}{r} \)).

Answer 1

Step 1: Let the two charges be at points \( A \) and \( B \), separated by distance \( 2a \). The point \( P \) is at distance \( r \) from the center \( O \).
Step 2: The distances from \( P \) to the charges are: \[ r_+ = r - a \cos \theta, \quad r_- = r + a \cos \theta, \] assuming \( r \gg a \).
Step 3: The potential at \( P \) is: \[ V = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{r_+} - \frac{q}{r_-} \right). \]
Step 4: Using binomial approximation for \( r \gg a \): \[ \frac{1}{r_\pm} \approx \frac{1}{r} \mp \frac{a \cos \theta}{r^2}. \]
Step 5: Substitute to get: \[ V = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r} - \frac{a \cos \theta}{r^2} - \frac{1}{r} - \frac{a \cos \theta}{r^2} \right) = \frac{1}{4 \pi \epsilon_0} \frac{2 q a \cos \theta}{r^2}. \]
Step 6: Since \( p = 2 q a \), the expression becomes: \[ \boxed{V = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2}}. \]