Practice question
Question
When a charged conductor is placed in contact with an uncharged conductor of smaller size, why does the
smaller conductor acquire a higher surface charge density upon reaching equilibrium?
Explanation
**Equipotential surface** is surface where V constant, no work done moving charge along it because W = q ΔV =0 when ΔV=0. Electric field E always perpendicular to equipotential surface, direction from higher to lower potential, magnitude E = -dV/dr, steeper potential gradient means stronger field. When two conductors reach equilibrium, they share charge and attain the same potential. Potential on a conductor's surface is V = (Q/4 π ε₀ R) for a sphere (or similar for other shapes). For equal V , (Q₁/R₁) = (Q₂/R₂) , so Q ∝ R . Surface charge density sigma = (Q/4 π R²) , so sigma ∝ (Q/R²)
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