Q.1 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 1 Solution

Topic: Electrostatic Potential and Capacitance · Official NTA Booklet Code M1

Question

Two charged spherical conductors of radius R1 and R2 are connected by a wire. Then the ratio of surface charge densities of the spheres (σ1/σ2) is:

✓

Verified Against NTA Final Answer Key

Option (2) : R2 / R1

📚 Electrostatic Potential and Capacitance Easy

📋 Given Parameters & Problem Setup

Radius of Sphere 1 R_1

Radius of Sphere 2 R_2

Condition Connected by a conducting wire (common potential V)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Potential Equality Shortcut: σ ∝ 1/R

⏱️ Save ~25 seconds

For any spherical conductor at potential V, $V = \frac{\sigma R}{\varepsilon_0} = \text{constant}$ . Therefore, surface charge density is inversely proportional to radius: $\sigma \propto \frac{1}{R} \implies \frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}$ ! Instant 5-second answer matching Option (2)!

1 When connected by a wire, charge redistributes until electric potentials equalize:

V_1 = V_2

2 Express potential in terms of surface charge density:

V = \frac{q}{4\pi\varepsilon_0 R} = \frac{\sigma (4\pi R^2)}{4\pi\varepsilon_0 R} = \frac{\sigma R}{\varepsilon_0}

3 Since

V_1 = V_2

, we have

\frac{\sigma_1 R_1}{\varepsilon_0} = \frac{\sigma_2 R_2}{\varepsilon_0}

4 Therefore,

\frac{\sigma_1}{\sigma_2} = \frac{R_2}{R_1}

5 Match Option (2).

🛡️

Why You Can 100% Trust This Shortcut: Canonical Gauss's law and electrostatic equilibrium of connected conductors.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Condition for Electrostatic Equilibrium of Connected Spheres

When two conducting spheres of radii $R_1$ and $R_2$ are joined by a thin conducting wire, free electrons flow between them until they reach an identical electrostatic potential $V$ :

V_1 = V_2 \implies \frac{1}{4\pi\varepsilon_0} \frac{q_1}{R_1} = \frac{1}{4\pi\varepsilon_0} \frac{q_2}{R_2} \implies \frac{q_1}{q_2} = \frac{R_1}{R_2}

Step 2

Relating Charge to Surface Charge Density

Surface charge density $\sigma$ is defined as the total charge per unit spherical surface area $A = 4\pi R^2$ :

\sigma_1 = \frac{q_1}{4\pi R_1^2}, \quad \sigma_2 = \frac{q_2}{4\pi R_2^2}

Step 3

Computing the Ratio of Surface Charge Densities

Substitute the charge ratio $q_1 / q_2 = R_1 / R_2$ into the density ratio:

\frac{\sigma_1}{\sigma_2} = \left(\frac{q_1}{q_2}\right) \times \left(\frac{R_2^2}{R_1^2}\right) = \left(\frac{R_1}{R_2}\right) \times \left(\frac{R_2^2}{R_1^2}\right) = \frac{R_2}{R_1}

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360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 1

✗ Incorrect

Verdict & Scientific Reason:

This represents the ratio of charges ( $q_1/q_2$ ), not the ratio of surface charge densities ( $sigma_1/sigma_2$ ).

⚠️

Student Fallacy (Why candidates pick this trap): Confusing charge ratio with surface charge density ratio.

2 2

✓ 100% Correct

Verdict & Scientific Reason:

Exact mathematical outcome derived from $V_1 = V_2 implies sigma_1 R_1 = sigma_2 R_2 implies sigma_1/sigma_2 = R_2/R_1$ .

💡

Why this is airtight: None; rigorous electrostatics derivation.

3 3

✗ Incorrect

Verdict & Scientific Reason:

Square-root dependence is unphysical for potential-charge relationships.

⚠️

Student Fallacy (Why candidates pick this trap): Erroneously introducing square roots from capacitive energy equations.

4 4

✗ Incorrect

Verdict & Scientific Reason:

This would occur if the spheres had identical charges ( $q_1 = q_2$ ), which is false for connected conductors of unequal radii.

⚠️

Student Fallacy (Why candidates pick this trap): Assuming charge remains unchanged instead of redistributing to equal potentials.

⚠️

NTA Trap Alert: Equal Charge vs. Equal Potential Trap

Students frequently assume connecting two spheres distributes charge equally ( $q_1 = q_2$ ). Connected conductors always equalize electric POTENTIAL ( $V_1 = V_2$ ), which means the smaller sphere has higher surface charge density (sharper curvature = greater $\sigma$ ).

🛡️ How to avoid the -1 mark: Remember: Connected by wire

\implies

equal POTENTIAL (

V

), NOT equal charge (

q

\sigma \propto 1/R

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Core NEET Syllabus Standard

NCERT Physics · Electrostatic Potential and Capacitance Standard Benchmark

Core Scientific Principle Tested:

"On a connected conducting surface in electrostatic equilibrium, electric potential is constant, and surface charge density is inversely proportional to the radius of curvature."

Syllabus Focus: Potential Due to a System of Charges and Conductors

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 1 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Potential Equality Shortcut: σ ∝ 1/R

Step-by-Step Rigorous Workflow & Mathematical Proof

Condition for Electrostatic Equilibrium of Connected Spheres

Relating Charge to Surface Charge Density

Computing the Ratio of Surface Charge Densities

360° Option Elimination Engine Key Feature

NTA Trap Alert: Equal Charge vs. Equal Potential Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — Electrostatic Potential and Capacitance

Electrostatic Energy Ratio in a Series-Parallel Capacitor Network

Path-Independence of Electrostatic Work & Potential of a Uniformly Charged Sphere

Electrostatic Shielding and Cavity Field with Interior Point Charge

Speed of an Attracted Point Charge Reaching the Surface of a Charged Insulating Sphere

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