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The axial electric field of a charged ring of radius R carrying charge Q at distance x from its centre is E = (1/(4*pi*e0)) * Qx/(R² + x²)^(3/2). At what distance from the centre is E maximum?

1. x = R
2. x = R/2
3. x = R/sqrt(2)
4. x = sqrt(2)*R

Correct answer: x = R/sqrt(2)

Solution

Maximizing E means maximizing f(x) = x/(R² + x²)^(3/2). Differentiating: f'(x) = [(R² + x²)^(3/2) - x*(3/2)(R²+x²)^(1/2)*2x] / (R²+x²)³. Setting numerator zero gives (R² + x²) - 3x² = 0, i.e. R² = 2x², so x = R/sqrt(2).

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