Question

The potential at a point at a distance \( r \) from the centre of an electric dipole is proportional to

• A. \( \frac{1}{r} \)
• B. \( \frac{1}{r^2} \)
• C. \( \frac{1}{r^3} \)
• D. \( r^2 \)

Hint:

Remember: Dipole potential $\longrightarrow$ \(1/r^2\), Dipole field $\longrightarrow$ \(1/r^3\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An electric dipole consists of two opposite charges. Unlike a single point charge whose potential decreases as \( 1/r \), the overlapping fields of two opposite charges in a dipole cause the potential to drop off more sharply with distance.
Step 2: Key Formula or Approach:
The electric potential \( V \) due to a dipole of moment \( p \) at a distance \( r \) is given by:
\[ V = \frac{1}{4\pi\epsilon_0} \frac{p \cos \theta}{r^2} \]
where \( \theta \) is the angle between the position vector and the dipole axis.
Step 3: Detailed Explanation:
Looking at the formula:
\[ V = \frac{k p \cos \theta}{r^2} \]
For a specific direction (fixed \( \theta \)), we can see that:
\[ V \propto \frac{1}{r^2} \]
This means if you double the distance from a dipole, the potential becomes one-fourth of its original value. In contrast, for a single point charge, it would only reduce to half.
Step 4: Final Answer:
The potential is proportional to \( 1/r^2 \).