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 P of magnitude, 4 x 10^-6 C m, is ±9 x 10^3 V. (Take 1/(4πε₀) = 9 x 10^9 SI units) Reason R : V = ± 2P/(4πε₀r²), 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:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Physics Q9 – NEET-UG 2024

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Home Physics Electric Charges and FieldsQuestion #104

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 \times 10^{- 6}

C m, is

\pm 9 \times 10^{3}

V. (Take

\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9}

SI units) Reason R :

V = \pm \frac{2 P}{4 π ϵ_{0} r^{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:

Clue 1 of 3: Electric Potential of a Dipole

Recall the formula for electric potential at an axial point due to an electric dipole.

🧠 Full Solution Path

Step 1: Recall the formula for electric potential on the axial line of a dipole✦ Active

The electric potential $V$ at an axial point at a distance $r$ from the center of a short electric dipole with dipole moment $P$ is given by:

V = \frac{1}{4 π ϵ_{0}} \frac{P}{r^{2}}

The sign of the potential depends on the direction of the dipole moment vector relative to the axial point.

Step 2: Evaluate Assertion A○ Expand

Given values are $P = 4 \times 10^{- 6}$ C m, $r = 2$ m, and $\frac{1}{4 π ϵ_{0}} = 9 \times 10^{9}$ N m $^{2}$ C $^{- 2}$ . Substitute these values into the formula for potential:

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

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

V = 9 \times 10^{9} \times 10^{- 6}

V = 9 \times 10^{3} V

Since the potential can be positive or negative depending on the orientation, $\pm 9 \times 10^{3}$ V is correct. Thus, Assertion A is true.

💡 Teacher's Secret Hint

Remember that potential is a scalar quantity, but its sign depends on the relative orientation of the dipole moment and the position vector.

Step 3: Evaluate Reason R and conclude○ Expand

Reason R states the formula for potential as $V = \pm \frac{2 P}{4 π ϵ_{0} r^{2}}$ . Comparing this with the correct formula for electric potential at an axial point, $V = \pm \frac{P}{4 π ϵ_{0} r^{2}}$ , we see that Reason R's formula has an extra factor of 2 in the numerator, which is incorrect. This formula is not for the potential on the axial line of a dipole. Therefore, Reason R is false.

Based on the evaluation, Assertion A is true, but Reason R is false. This corresponds to option (2).

💡 Teacher's Secret Hint

Be careful not to confuse the formula for electric potential with that for electric field, or with potential at an equatorial point.

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Physics Question 9 – NEET-UG 2024

Clue 1 of 3: Electric Potential of a Dipole

🧠 Full Solution Path