Surface Charge Density After Contact – Ratio σ₁/σ₂ | JEE Electrostatics | Doubtify JEE

 

⚡ Surface Charge Density After Contact – Electrostatics | JEE Physics | Doubtify JEE


📌 Question:

Two metal spheres of radius R and 3R have the same surface charge density σ.
They are brought in contact and then separated.
After separation, let the surface charge densities be σ₁ (for smaller sphere) and σ₂ (for larger sphere).
What is the ratio σ₁ / σ₂?


🖼️ Question Image:

Two metal spheres of radius R and 3R have the same surface charge density σ.
🧠 Solution Image:

🧪 Detailed Concept & Solution:

⚙️ Given:

  • Radii: R (small sphere) and 3R (large sphere)

  • Initial surface charge density: σ on both spheres
    ⇒ So initial charges:
    Q1=σ4πR2Q_1 = \sigma \cdot 4\pi R^2
    Q2=σ4π(3R)2=9σ4πR2Q_2 = \sigma \cdot 4\pi (3R)^2 = 9\sigma \cdot 4\pi R^2

🔁 When they are brought in contact:

  • Charge redistributes such that potential becomes equal
    For a sphere,
    V=14πε0QRV = \dfrac{1}{4\pi \varepsilon_0} \cdot \dfrac{Q}{R}

Let final charges be Q₁′ and Q₂′ after contact.

So:

Q1R=Q23RQ2=3Q1\frac{Q_1'}{R} = \frac{Q_2'}{3R} \Rightarrow Q_2' = 3Q_1'

Also, total charge is conserved:

Q1+Q2=Q1+Q2=σ4πR2(1+9)=10σ4πR2Q_1' + Q_2' = Q_1 + Q_2 = \sigma \cdot 4\pi R^2 (1 + 9) = 10\sigma \cdot 4\pi R^2

Substitute Q₂′ = 3Q₁′:

Q1+3Q1=4Q1=10σ4πR2Q1=2.5σ4πR2Q_1' + 3Q_1' = 4Q_1' = 10\sigma \cdot 4\pi R^2 \Rightarrow Q_1' = 2.5\sigma \cdot 4\pi R^2

So:

Q2=7.5σ4πR2Q_2' = 7.5\sigma \cdot 4\pi R^2

✅ Now, Surface Charge Density:

σ1=Q14πR2=2.5σσ2=Q24π(3R)2=7.5σ4πR24π9R2=7.5σ9σ_1 = \frac{Q_1'}{4\pi R^2} = 2.5σ \quad σ_2 = \frac{Q_2'}{4\pi (3R)^2} = \frac{7.5σ \cdot 4\pi R^2}{4\pi 9R^2} = \frac{7.5σ}{9}

So:

σ1σ2=2.5σ7.5σ9=2.597.5=22.57.5=3\frac{σ_1}{σ_2} = \frac{2.5σ}{\frac{7.5σ}{9}} = \frac{2.5 \cdot 9}{7.5} = \frac{22.5}{7.5} = 3

✅ Final Answer:

σ1/σ2=3\boxed{σ_1 / σ_2 = 3}

🎥 Video Solution:

🔍 Why This Question is Important:

  • Conceptual mix of Electrostatics + Charge Sharing

  • Tests your understanding of potential equalization and surface area-based reasoning

  • Commonly asked in JEE Main and Advanced

  • Mastering this helps in solving problems involving charge redistribution in conductors


🧠 Pro Tip:

Whenever two conducting spheres are connected, their potentials equalize — not charge or surface density. Always start from that principle.


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