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A 6 cm long uniformly charged plastic rod (total charge Q = 10 nC) is placed near a long wire with linear charge density lambda = 10⁻⁷ C/m. The nearest end of the rod is 2 cm from the wire. Calculate the electric force on the rod. Express the answer as alpha * 10⁻⁴ N and find alpha.
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Correct answer: 2
Solution
The force on a small element dr of the rod at distance r from wire is dF = (Q/L) * dr * lambda/(2*pi*epsilon₀*r). Total force F = (Q*lambda)/(2*pi*epsilon₀*L) * integral[0.02 to 0.08] dr/r = (Q*lambda)/(2*pi*epsilon₀*L) * ln(0.08/0.02) = (Q*lambda)/(2*pi*epsilon₀*L) * ln(4). Substituting: Q = 10e-9 C, lambda = 10⁻⁷ C/m, L = 0.06 m, 1/(2*pi*epsilon₀) = 2*9*10⁹ (wait, = 2*k where k=9e9... no: E = lambda/(2*pi*eps0*r) = 2k*lambda/r). F = 2k * Q * lambda * ln(4) / L = 2 * 9e9 * 10e-9 * 1e-7 * ln(4) / 0.06 = 2 * 9e9 * 10e-9 * 1e-7 * 1.386 / 0.06 = 2 * 9 * 10 * 1e-7 * 1.386/0.06 = 2 * 9 * 10 * 2.31e-6 = 2 * 2.079e-4 = 4.158e-4 N. Hmm that gives alpha = 4. Let me redo: F = (lambda_wire * Q) / (2*pi*eps0 * L) * ln(r2/r1). = (9e9 * 2 * lambda_wire * Q / L) * ln(4). = (2 * 9e9 * 1e-7 * 10e-9 / 0.06) * ln(4). = (2 * 9e9 * 1e-7 * 1e-8 / 0.06) * 1.386. = (2 * 9e9 * 1e-15 / 0.06) * 1.386. = (2 * 9e9 * 1.667e-14) * 1.386. = (2 * 1.5e-4) * 1.386 = 3e-4 * 1.386 = 4.16e-4 N. So alpha approx 4. But the answer might be 2 if ln(4) approximated differently or if I recalculate. Let me verify: 2*k = 2*9*10⁹ = 1.8*10¹⁰. lambda_wire = 10⁻⁷ C/m. Q_rod = 10⁻⁸ C. L_rod = 0.06 m. r1 = 0.02 m, r2 = 0.08 m. F = (2k*lambda*Q/L)*ln(r2/r1) = (1.8e10 * 1e-7 * 1e-8/0.06)*ln(4) = (1.8e10 * 1.667e-16)*1.386 = (3e-6)*1.386 = 4.16e-6 N = 0.416*10⁻⁵ N. That gives alpha much less than 1. Let me recheck: Q = 10 nC = 10*10⁻⁹ = 10⁻⁸ C. k = 9*10⁹. F = (2k*lambda*Q/L)*ln(4) = 2*9e9*(1e-7)*(1e-8)/(0.06)*ln4 = (18e9*1e-15/0.06)*1.386 = (18e-6/0.06)*1.386 = 3e-4*1.386 = 4.16e-4 N. So alpha = 4.16 approx 4. The answer seems to be 4.
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