Question

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

• A. 1/r² and 1/r³
• B. 1/r and 1/r²
• C. 1/r³ and 1/r²
• D. 1/r² and 1/r

Answer 1

The Correct Option is C

Electric Field:

\( E_g = \frac{1}{4 \pi \epsilon_0} \frac{p}{r^3} \sqrt{3 \cos^2 \theta + 1} \)

Electric Potential: 

\( V_g = \frac{1}{4 \pi \epsilon_0} \frac{p \cos \theta}{r^2} \)

Therefore, the correct option is: D.

Answer 2

Given:

\( E = K r^{3/2} \quad \text{and} \quad E \propto r^{3/1} \) 

Similarly, for the potential \( V \), we have:

\( V = K r^2 p \cos \theta \quad \text{and} \quad V \propto r^2 \)

Thus, the relationships show that:

• Energy \( E \) is proportional to \( r^{3/1} \)
• Potential \( V \) is proportional to \( r^2 \)

Answer 3

We need to determine how the electric potential (\( V \)) and the electric field (\( E \)) due to an electric dipole depend on the distance (\( r \)) from the dipole, for distances much larger than the separation of the charges in the dipole (\( r \gg d \)).

Electric Potential (\( V \)) of a Dipole:
The electric potential at a point \( P \) located at a distance \( r \) from the center of the dipole and at an angle \( \theta \) with respect to the dipole axis is given by: \[ V = \frac{1}{4\pi\epsilon_0} \frac{p \cos\theta}{r^2} \] where \( p \) is the magnitude of the electric dipole moment. From this expression, we can see that the electric potential \( V \) is proportional to \( \frac{1}{r^2} \). \[ \mathbf{V \propto \frac{1}{r^2}} \]

Electric Field (\( E \)) of a Dipole:
The electric field can be derived from the potential using \( \vec{E} = -\nabla V \). In polar coordinates, the radial (\( E_r \)) and tangential (\( E_\theta \)) components of the electric field are: \[ E_r = -\frac{\partial V}{\partial r} = -\frac{\partial}{\partial r} \left( \frac{p \cos\theta}{4\pi\epsilon_0 r^2} \right) = \frac{p \cos\theta}{4\pi\epsilon_0} \left( \frac{2}{r^3} \right) = \frac{1}{4\pi\epsilon_0} \frac{2p \cos\theta}{r^3} \] \[ E_\theta = -\frac{1}{r}\frac{\partial V}{\partial \theta} = -\frac{1}{r} \frac{\partial}{\partial \theta} \left( \frac{p \cos\theta}{4\pi\epsilon_0 r^2} \right) = -\frac{p}{4\pi\epsilon_0 r^3} (-\sin\theta) = \frac{1}{4\pi\epsilon_0} \frac{p \sin\theta}{r^3} \] The magnitude of the total electric field \( E \) is given by \( E = \sqrt{E_r^2 + E_\theta^2} \): \[ E = \sqrt{\left(\frac{1}{4\pi\epsilon_0} \frac{2p \cos\theta}{r^3}\right)^2 + \left(\frac{1}{4\pi\epsilon_0} \frac{p \sin\theta}{r^3}\right)^2} \] \[ E = \frac{p}{4\pi\epsilon_0 r^3} \sqrt{(2\cos\theta)^2 + (\sin\theta)^2} \] \[ E = \frac{p}{4\pi\epsilon_0 r^3} \sqrt{4\cos^2\theta + \sin^2\theta} \] \[ E = \frac{p}{4\pi\epsilon_0 r^3} \sqrt{3\cos^2\theta + \cos^2\theta + \sin^2\theta} \] \[ E = \frac{p}{4\pi\epsilon_0 r^3} \sqrt{3\cos^2\theta + 1} \] From this expression, we see that the magnitude of the electric field \( E \) is proportional to \( \frac{1}{r^3} \). \[ \mathbf{E \propto \frac{1}{r^3}} \]

Therefore, the electric field \( E \) varies as \( \mathbf{1/r^3} \) and the electric potential \( V \) varies as \( \mathbf{1/r^2} \).

Comparing this with the given options (order: E variation, V variation):

• 1/r² and 1/r³
• 1/r and 1/r²
• 1/r³ and 1/r²
• 1/r² and 1/r

The correct option is the third one.