Question:

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

Updated On: Apr 1, 2025
  • 1/r2 and 1/r3

  • 1/r and 1/r2

  • 1/r3 and 1/r2

  • 1/r2 and 1/r

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The Correct Option is C

Approach Solution - 1

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.

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Approach Solution -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 \( r^2 \)
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Top Questions on Electromagnetic induction

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Concepts Used:

Electromagnetic Induction

Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-

  1. When we place the conductor in a changing magnetic field.
  2. When the conductor constantly moves in a stationary field.

Formula:

The electromagnetic induction is mathematically represented as:-

e=N × d∅.dt

Where

  • e = induced voltage
  • N = number of turns in the coil
  • Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
  • t = time

Applications of Electromagnetic Induction

  1. Electromagnetic induction in AC generator
  2. Electrical Transformers
  3. Magnetic Flow Meter