Two positive point charges 5.00 ?C are fixed on the y-axis at y = 4.25 mm and y

Question

Two positive point charges 5.00 ?C are fixed on the y-axis at y = 4.25 mm and y = -4.25 mm. A negative point charge -4.25 ?C is located 9.5 mm on the x-axis. (a) What is the x-component of the force on the -4.25 ?C charge? (b) What is the y-component of the force on the -4.25 ?C charge? (c) What is the magnitude of the net force on the -4.25 ?C charge if it is located at the origin? (d) Where must the -4.25 ?C charge be place to obtain the maximum force on it from the 2 charges q? (e) What is that maximum force (magnitude only)?

Explanation / Answer

A) Use Coulomb's law: F = kq1q2 / r² where q1 and q2 are the charges of the two point charges in question, and r is the distance between the two. k is a proportionality constant given by: k = 1 / 4pe0 ˜ 8.99 × 10? Nm²C?² In your case: q1 = Q1 = 3.10 µC = 3.10 × 10?6 C q2 = Q3 = 6.30 µC = 6.30 × 10?6 C r = v(0.05² + 0.04²) = v0.0041 so the magnitude of the electric force on Q1 due to Q3 is given by: F = (8.99 × 10?)(3.10 × 10?6)(6.30 × 10?6) / (v0.0041)² = 42.82 N B) Force on Q1 due to Q2 = (8.99 × 10?)(3.10 × 10?6)(5.20 × 10?6) / (0.02² + 0.02²) = 181.15 N û where û = (1/v2)[1, -1] û is the unit vector pointing in the direction of the force! A unit vector is a vector with length = 1, that is why i have divided [1, -1] by v2. The angle between û and the negative x-axis is 45°, so the y component of the force is; sin(45°)(181.15) = (1/v2)(181.15) N = 128.09 N

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