Physics Q38 – JEE-MAIN 2025

Home Physics Electric Charges and FieldsQuestion #120

Two metal spheres of radius

R

and

3 R

have same surface charge density

σ

. If they are brought in contact and then separated, the surface charge density on smaller and bigger sphere becomes

σ_{1}

and

σ_{2}

, respectively. The ratio

\frac{σ_{1}}{σ_{2}}

Clue 1 of 3: Electrostatic Equilibrium

When conducting spheres are brought into contact, charge redistributes until their electrostatic potentials become equal.

🥷

Ninja Strategy•Potential Equalization Principle

When conductors are in contact, their potentials become equal, leading to the relationship $Q \propto R$ and consequently $σ \propto 1 / R$ . This means the ratio $σ_{1} / σ_{2} = R_{2} / R_{1}$ .

🧠 Full Solution Path

Step 1: Initial Charges and Total Charge✦ Active

Let the radii of the two spheres be $R_{1} = R$ and $R_{2} = 3 R$ . Initially, both spheres have the same surface charge density $σ$ . The initial charges are:

Q_{1} = σ \cdot 4 π R_{1}^{2} = σ \cdot 4 π R^{2}

Q_{2} = σ \cdot 4 π R_{2}^{2} = σ \cdot 4 π (3 R)^{2} = 9 σ \cdot 4 π R^{2}

The total initial charge is:

Q_{t o t a l} = Q_{1} + Q_{2} = σ \cdot 4 π R^{2} + 9 σ \cdot 4 π R^{2} = 10 σ \cdot 4 π R^{2}

Step 2: Charge Redistribution and Final Potentials○ Expand

When the two conducting spheres are brought into contact, charge redistributes until they reach the same electrostatic potential. Let the final charges be $Q_{1}^{'}$ and $Q_{2}^{'}$ . The final potentials are equal:

V_{1}^{'} = V_{2}^{'}

Using the formula for potential of a sphere $V = \frac{k Q}{R}$ :

\frac{k Q_{1}^{'}}{R_{1}} = \frac{k Q_{2}^{'}}{R_{2}} ⟹ \frac{Q_{1}^{'}}{R} = \frac{Q_{2}^{'}}{3 R} ⟹ Q_{2}^{'} = 3 Q_{1}^{'}

The total charge is conserved, so $Q_{1}^{'} + Q_{2}^{'} = Q_{t o t a l}$ . Substituting $Q_{2}^{'} = 3 Q_{1}^{'}$ :

Q_{1}^{'} + 3 Q_{1}^{'} = Q_{t o t a l} ⟹ 4 Q_{1}^{'} = Q_{t o t a l}

Thus, $Q_{1}^{'} = \frac{Q_{t o t a l}}{4}$ and $Q_{2}^{'} = \frac{3 Q_{t o t a l}}{4}$ .

💡 Teacher's Secret Hint

Remember that charge conservation and potential equalization are key principles for conductors in contact.

Step 3: Final Surface Charge Densities and Ratio○ Expand

The final surface charge densities are $σ_{1} = \frac{Q_{1}^{'}}{4 π R_{1}^{2}}$ and $σ_{2} = \frac{Q_{2}^{'}}{4 π R_{2}^{2}}$ . We need to find the ratio $\frac{σ_{1}}{σ_{2}}$ :

\frac{σ_{1}}{σ_{2}} = \frac{Q_{1}^{'} / (4 π R_{1}^{2})}{Q_{2}^{'} / (4 π R_{2}^{2})} = \frac{Q_{1}^{'}}{Q_{2}^{'}} \cdot \frac{R_{2}^{2}}{R_{1}^{2}}

From Step 2, we know that $\frac{Q_{1}^{'}}{Q_{2}^{'}} = \frac{R_{1}}{R_{2}}$ . Substituting this into the ratio:

\frac{σ_{1}}{σ_{2}} = (\frac{R_{1}}{R_{2}}) \cdot \frac{R_{2}^{2}}{R_{1}^{2}} = \frac{R_{2}}{R_{1}}

Given $R_{1} = R$ and $R_{2} = 3 R$ :

\frac{σ_{1}}{σ_{2}} = \frac{3 R}{R} = 3

💡 Teacher's Secret Hint

Notice the inverse relationship between surface charge density and radius for conductors in equilibrium: $σ \propto 1 / R$ .

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Physics Question 38 – JEE-MAIN 2025

Clue 1 of 3: Electrostatic Equilibrium

🧠 Full Solution Path