Question

Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R.
Assertion A : The potential (V) at any axial point, at 2 m distance(r) from the centre of the dipole of dipole moment vector

 \(\vec{P}\) of magnitude, 4 × 10^-6 C m, is ± 9 × 10³ V.
      (Take \(\frac{1}{4\pi\epsilon_0}=9\times10^9\) SI units)
 

Reason R : \(V=±\frac{2P}{4\pi \epsilon_0r^2}\), where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below :

• A. Both A and R are true and R is the correct explanation of A.
• B. Both A and R are true and R is NOT the correct explanation of A.
• C. A is true but R is false.
• D. A is false but R is true.

Answer 1

The Correct Option is C

Step 1: Calculate the Potential on the Axial Line of a Dipole

The potential at an axial point of a dipole is given by the formula:

$$ V = \frac{P}{4\pi\epsilon_0 r^2} $$

Where:

\( V \) represents the Potential.

\( P \) represents the Dipole moment.

\( r \) represents the Distance from the dipole center.

\( \epsilon_0 \) represents the Permittivity of free space.

Step 2: Substitute the Given Values

The provided values are:

• \( P = 4 \times 10^{-6} \) Cm
• \( r = 2 \) m
• \( \frac{1}{4\pi\epsilon_0} = 9 \times 10^9 \) SI units

Upon substitution:

$$ V = \frac{(4 \times 10^{-6}) \cdot (9 \times 10^9)}{2^2} $$

The solution yields:

$$ V = \frac{36 \times 10^3}{4} = 9 \times 10^3 \text{ V} $$

This result aligns with the assertion.

Step 3: Verify the Reason

The provided reason states the formula:

$$ V = \pm \frac{2P}{4\pi\epsilon_0 r^2} $$

This formula is accurate; the ± sign indicates the potential's direction relative to the dipole's orientation.

However, the assertion calculated the potential's magnitude, which is independent of this directional sign.

Conclusion

Consequently, the reason is not applicable to the assertion.