Question

Which one is the correct expression of electric potential on the axial point of a dipole whose dipole moment is р and length is 2l, at a distance of r?

• A. \(\frac{p}{r^2}\)
• B. \(\frac{p}{2lr}\)
• C. \(\frac{p}{(r^2-l^2)^2}\)
• D. \(\frac{p}{l^2}\)

Answer 1

The Correct Option is A

We need to find the correct expression for the electric potential \( V \) at a point on the axial line of an electric dipole with dipole moment \( \mathbf{p} \) and length \( 2l \), at a distance \( r \) from the dipole’s center, and select the correct option from: \( \frac{p}{r^2} \), \( \frac{p}{2lr} \), \( \frac{p}{(r^2 - l^2)^2} \), and \( \frac{p}{l^2} \).

1. Electric Dipole Configuration:
An electric dipole consists of two charges, \( +q \) and \( -q \), separated by a distance \( 2l \). The dipole moment is \( \mathbf{p} = q \cdot 2l \), directed from the negative to the positive charge. The axial point lies along the dipole axis at a distance \( r \) from the center of the dipole.

2. Electric Potential Formula:
The electric potential due to a point charge \( q \) at distance \( d \) is:

\( V = \frac{1}{4 \pi \epsilon_0} \frac{q}{d} \)
where \( \epsilon_0 \) is the permittivity of free space. The total potential is the sum of contributions from both charges.

3. Positioning the Charges:
Place the dipole along the x-axis, centered at the origin, with \( -q \) at \( x = -l \) and \( +q \) at \( x = +l \). For a point on the axial line at \( x = r \) (where \( r > l \)), calculate the distances:
- Distance from \( +q \) (at \( x = l \)) to the point: \( r - l \).
- Distance from \( -q \) (at \( x = -l \)) to the point: \( r - (-l) = r + l \).

4. Calculating the Potential:
The potential due to the positive charge:

\( V_+ = \frac{1}{4 \pi \epsilon_0} \frac{q}{r - l} \)
The potential due to the negative charge:

\( V_- = \frac{1}{4 \pi \epsilon_0} \frac{-q}{r + l} \)
The total potential is:

\( V = V_+ + V_- = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{r - l} - \frac{q}{r + l} \right) \)

5. Simplifying the Expression:
Factor out common terms:

\( V = \frac{q}{4 \pi \epsilon_0} \left( \frac{1}{r - l} - \frac{1}{r + l} \right) \)
Combine the fractions:

\( \frac{1}{r - l} - \frac{1}{r + l} = \frac{(r + l) - (r - l)}{(r - l)(r + l)} = \frac{r + l - r + l}{r^2 - l^2} = \frac{2l}{r^2 - l^2} \)
Thus:

\( V = \frac{q}{4 \pi \epsilon_0} \cdot \frac{2l}{r^2 - l^2} = \frac{1}{4 \pi \epsilon_0} \frac{q \cdot 2l}{r^2 - l^2} \)
Since \( p = q \cdot 2l \), we get:

\( V = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2 - l^2} \)

6. Approximation for Large \( r \):
In many physics problems, \( r \gg l \), so \( l^2 \) is negligible compared to \( r^2 \):

\( r^2 - l^2 \approx r^2 \)
Thus:

\( V \approx \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2} \)

7. Comparing with Given Options:
The exact expression is \( V = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2 - l^2} \), but none of the options include the factor \( \frac{1}{4 \pi \epsilon_0} \), suggesting the problem expects the potential without this constant (common in some contexts where constants are omitted for simplicity) or assumes the far-field approximation. Let’s evaluate the options:
- \( \frac{p}{r^2} \): Matches the approximate form for \( r \gg l \).
- \( \frac{p}{2lr} \): Does not match dimensionally or mathematically, as it gives units of \( \frac{\text{C} \cdot \text{m}}{\text{m} \cdot \text{m}} = \frac{\text{C}}{\text{m}} \), incorrect for potential.
- \( \frac{p}{(r^2 - l^2)^2} \): Matches the denominator structure but has a squared term, leading to incorrect units (\( \frac{\text{C} \cdot \text{m}}{\text{m}^4} \)) and magnitude.
- \( \frac{p}{l^2} \): Yields units of \( \frac{\text{C} \cdot \text{m}}{\text{m}^2} = \frac{\text{C}}{\text{m}} \), incorrect for potential, and lacks dependence on \( r \).

8. Selecting the Correct Option:
The option \( \frac{p}{r^2} \) corresponds to the standard far-field approximation \( V \approx \frac{1}{4 \pi \epsilon_0} \frac{p}{r^2} \) (without the constant, as per the options provided). The exact form \( \frac{p}{r^2 - l^2} \) is not listed, and the other options are dimensionally or mathematically inconsistent.

Final Answer:
The correct expression for the electric potential on the axial point of the dipole, assuming the far-field approximation (\( r \gg l \)), is: \( \frac{p}{r^2} \)