Electrostatics: Edge Cases and Subtle Traps (3)

hard • 2 min read
Tags Electrostatics

Question

Two identical metallic spheres carry charges +6 μC+6\,\mu\text{C} and −2 μC-2\,\mu\text{C}. They are brought into contact and then separated to their original distance of 0.1 m0.1\text{ m}. (a) Find the new charge on each sphere. (b) Find the force between them after separation. (c) Compare with the original force before contact.

Solution — Step by Step

When two identical conductors touch, charge redistributes to make their potentials equal. For identical spheres, this means equal charges on each. Total charge is conserved.

Qeach=+6+(−2)2=+2 μCQ_{\text{each}} = \frac{+6 + (-2)}{2} = +2\,\mu\text{C}

Both spheres end up with +2 μC+2\,\mu\text{C}.

Now both charges are positive, so the force is repulsive. Using Coulomb’s law:

Fafter=14πε0⋅Q1Q2r2=9×109×(2×10−6)2(0.1)2F_{\text{after}} = \frac{1}{4\pi\varepsilon_0}\cdot\frac{Q_1 Q_2}{r^2} = \frac{9\times10^9 \times (2\times10^{-6})^2}{(0.1)^2} Fafter=9×109×4×10−120.01=3.6 N (repulsive)F_{\text{after}} = \frac{9\times10^9 \times 4\times10^{-12}}{0.01} = 3.6\text{ N (repulsive)}

Before contact, charges were +6 μC+6\,\mu\text{C} and −2 μC-2\,\mu\text{C} — opposite signs, so attractive.

Fbefore=9×109×6×10−6×2×10−6(0.1)2=10.8 N (attractive)F_{\text{before}} = \frac{9\times10^9 \times 6\times10^{-6} \times 2\times10^{-6}}{(0.1)^2} = 10.8\text{ N (attractive)}

After contact: +2 μC+2\,\mu\text{C} each, force =3.6 N= 3.6\text{ N} repulsive. Before contact: 10.8 N10.8\text{ N} attractive.

Why This Works

The “identical conductors share equally” rule is a consequence of equal capacitance — same geometry means same CC, and at equilibrium the potential is uniform across the connected system, so Q=CVQ = CV gives equal QQ.

Coulomb’s law uses the product Q1Q2Q_1 Q_2 with sign. Positive product means repulsive, negative means attractive. The magnitude tracks the absolute product.

Alternative Method

For non-identical spheres, the redistribution is in the ratio of their radii: Q1′/Q2′=r1/r2Q_1' / Q_2' = r_1 / r_2 (since potentials must equalise). This is a JEE-Advanced favourite — they swap “identical” for “radii in ratio 2:1” and watch students mechanically halve the total.

Common Mistake

Forgetting the sign change in force direction. The force was attractive before contact, repulsive after. If a question asks for “change in force,” you must use signed values: ΔF=(−3.6)−(+10.8)=−14.4 N\Delta F = (-3.6) - (+10.8) = -14.4\text{ N} (taking attractive as positive). A blind subtraction 10.8−3.6=7.2 N10.8 - 3.6 = 7.2\text{ N} misses the physics entirely.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next

Capacitor combinations — series and parallel with charge and voltage distribution

medium

Capacitors in Series and Parallel — Equivalent Capacitance

easy

Coulomb's Law vs Gravitational Force — Why Electrostatic Is Stronger

easy

Derive expression for electric field due to infinite charged plane using Gauss's law

medium

Electric Dipole — Field on Axial and Equatorial Points

medium