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 $2 \mathrm{~m}$ distance $(r)$ from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 \times 10^{-6} \mathrm{~C} \mathrm{m},$ is $\pm 9 \times 10^3 \mathrm{~V}.$ $\text{(Take }\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{~SI ~units})$ Reason (R): $V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2},$ where $r$ is the distance of any axial point, situated at $2 \mathrm{~m}$ from the centre of the dipole. In the light of the above statements, choose the correct answer from the options given below:

## CORE INFORMATION\n✓ **Answer**: A\n## CONCEPT FRAMEWORK\n\t**Primary Concept**: Electric Potential due to a Dipole\n\t**Related Topics**: Electrostatics, Dipoles\n\t**Prerequisites**: Coulomb's Law, Electric Potential\n## SOLUTION PATHWAY\n**Given Information**:\n\t$P = 4 \times 10^{-6} \mathrm{~C} \mathrm{m}$\n\t$r = 2 \mathrm{~m}$\n\t$\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \mathrm{~SI ~units}$\n**Method & Steps**:\n\t1. Calculate the potential (V) using the correct formula.\n\t\tReasoning: The electric potential at an axial point due to a dipole is given by $V = \frac{P}{4 \pi \epsilon_0 r^2}$.\n\t\tFormula/Rule: $V = \frac{P}{4 \pi \epsilon_0 r^2}$\n\t2. Substitute the given values.\n\t\tWorking: $V = (9 \times 10^9) \times \frac{4 \times 10^{-6}}{(2)^2} = 9 \times 10^3 \mathrm{~V}$\n\t\tKey Point: The given assertion (A) uses $\pm$ which is correct as the potential can be positive or negative depending on which side of the dipole the point is located. The reason (R) also uses $\pm$ but is incorrect because of the factor '2' in the numerator and '$r^2$' in the denominator which should be 'r'.\n\t3. Compare the calculated value with the assertion.\n\t\tResult: The calculated value matches the assertion (A).\n\t\tVerification: The assertion is correct.\n ## OPTION ANALYSIS\n[A] ✓: (A) is correct as the calculated potential matches the given value. (R) is incorrect as the formula given in (R) has an extra factor of 2 in the numerator and r is squared in the denominator, the correct formula is $V = \pm \frac{P}{4 \pi \epsilon_0 r^2}$.\n[B] :x: (A) is correct, so this option is incorrect.\n[C] :x: (R) is not correct, so this option is incorrect.\n[D] :x: (R) is not correct, so this option is incorrect.\n ## LEARNING AIDS\n\t**Common Mistakes**:\n\t\t1.Incorrect Formula: Using $V = \pm \frac{2P}{4 \pi \epsilon_0 r^2}$ instead of $V = \pm \frac{P}{4 \pi \epsilon_0 r^2}$.\n\t\t\tHow to avoid: Carefully remember the correct formula for electric potential due to a dipole.\n\t\t2.Confusing Axial and Equatorial Points: The formulas for potential at axial and equatorial points are different. At the equatorial point, V=0.\n\t\t\tHow to avoid: Clearly identify whether the point is axial or equatorial before applying the formula.\n\t**Quick Tips**:\n\t\t1.Remember the proportionality: $V \propto \frac{1}{r^2}$ for an axial point and $V=0$ for an equatorial point.\n\t\t2.Visualize the dipole and the field lines to understand the potential variation.

The correct Option is "(A) is correct but (R) is not correct"

NEET Physics Electrostatic Potential And Capacitance MCQs

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NEET Questions / Physics / Electrostatic Potential And Capacitance

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Electric Potential And Potential Energy

Electric Potential Due To Dipole

Torque Acting On Dipole

Relation Between E and V

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Different Types Of Capacitor

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Capacitors In Series And In Parallel

Energy Stored In A Condenser

Electric Dipole

1681.

A capacitor has air as dielectric medium and two conducting plates of area $12 \mathrm{~cm}^2$ and they are $0.6 \mathrm{~cm}$ apart. When a slab of dielectric having area $12 \mathrm{~cm}^2$ and $0.6 \mathrm{~cm}$ thickness is inserted between the plates, one of the conducting plates has to be moved by $0.2 \mathrm{~cm}$ to keep the capacitance same as in previous case. The dielectric constant of the slab is: $\left(ext{Given }\varepsilon_0=8.834 imes 10^{-12} \mathrm{~F} / \mathrm{m}\right)$

1.5

1.33

0.66

Question Tags

📚 Subject:Physics

📑 Chapter:Electrostatic Potential And Capacitance

🎯Topic:Capacitance

🎓 Ncert Level:Application

🎯Difficulty :Easy

⚡ Time Required :3 mins

🔄Ncert Concepts :Capacitance, Dielectric Constant, Electric Fields

1682.

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 $2 \mathrm{~m}$ distance $(r)$ from the centre of the dipole of dipole moment vector $\vec{P}$ of magnitude, $4 imes 10^{-6} \mathrm{~C} \mathrm{m},$ is $\pm 9 imes 10^3 \mathrm{~V}.$

$ext{(Take }\frac{1}{4 \pi \epsilon_0}=9 imes 10^9 \mathrm{~SI ~units})$

Reason (R): $V= \pm \frac{2 P}{4 \pi \epsilon_0 r^2},$ where $r$ is the distance of any axial point, situated at $2 \mathrm{~m}$ from the centre of the dipole.