The electric field and the potential of an electric dipole vary with distance 'r' as

Options

A. `(1)/(r ) and (1)/(r^(2))`
B. `(1)/(r^(2)) and (1)/(r )`
C. `(1)/(r^(2)) and (1)/(r^(3))`
D. `(1)/(r^(3)) and (1)/(r^(2))`

Answer

To analyze how the electric field and potential of an electric dipole vary with distance \( r \), we can follow these steps:

### Step 1: Understanding the Electric Dipole
An electric dipole consists of two equal and opposite charges, \( +q \) and \( -q \), separated by a distance \( 2L \). The dipole moment \( P \) is defined as:
\[ P = q \cdot 2L \]

### Step 2: Electric Potential \( V \) at a Distance \( r \)
The electric potential \( V \) at a point located at a distance \( r \) from the center of the dipole (making an angle \( \theta \) with the dipole axis) is given by the formula:
\[ V = \frac{P \cos \theta}{4 \pi \epsilon_0 r^2} \]
where \( \epsilon_0 \) is the permittivity of free space.

### Step 3: Electric Field \( E \) from Potential
To find the electric field \( E \), we need to take the negative gradient of the potential. The electric field in the radial direction \( E_r \) can be derived as follows:
\[ E_r = -\frac{dV}{dr} \]
Substituting the expression for \( V \):
\[ E_r = -\frac{d}{dr} \left( \frac{P \cos \theta}{4 \pi \epsilon_0 r^2} \right) \]

### Step 4: Differentiate the Potential
Now, we differentiate \( V \) with respect to \( r \):
\[ E_r = -\left( \frac{P \cos \theta}{4 \pi \epsilon_0} \cdot \frac{d}{dr} \left( \frac{1}{r^2} \right) \right) \]
The derivative of \( \frac{1}{r^2} \) is:
\[ \frac{d}{dr} \left( \frac{1}{r^2} \right) = -\frac{2}{r^3} \]
Thus,
\[ E_r = -\left( \frac{P \cos \theta}{4 \pi \epsilon_0} \cdot -\frac{2}{r^3} \right) = \frac{2P \cos \theta}{4 \pi \epsilon_0 r^3} \]

### Step 5: Electric Field Components
The electric field has two components: the radial component \( E_r \) and the perpendicular component \( E_\theta \). The radial component is:
\[ E_r = \frac{2P \cos \theta}{4 \pi \epsilon_0 r^3} \]
The perpendicular component \( E_\theta \) can also be derived similarly, leading to:
\[ E_\theta = \frac{P \sin \theta}{4 \pi \epsilon_0 r^3} \]

### Conclusion
From the above derivations, we conclude that:
- The electric potential \( V \) varies as \( \frac{1}{r^2} \).
- The electric field \( E \) varies as \( \frac{1}{r^3} \).

### Summary of Results
- Electric Potential \( V \propto \frac{1}{r^2} \)
- Electric Field \( E \propto \frac{1}{r^3} \)