Question

If \(\mathbf{E}=120 \mathbf{a}_{\rho} \mathrm{V} / \mathrm{m}\), find the incremental amount of work done in moving a \(50-\mu \mathrm{C}\) charge a distance of \(2 \mathrm{~mm}\) from \((a) P(1,2,3)\) toward \(Q(2,1,4) ;(b)\) \(Q(2,1,4)\) toward \(P(1,2,3) .\)

Short answer

Solution: The incremental work done for (a) is: ΔW_a = (50 * 10^-6 C) * (120 V/m) * 0.002 * (1/sqrt(3)) The incremental work done for (b) is: ΔW_b = (50 * 10^-6 C) * (120 V/m) * 0.002 * (-1/sqrt(3))

Step-by-step solution

Calculate the displacement vector

To calculate the displacement vector ∆𝑑 between points P and Q, we simply subtract the coordinates of point P from the coordinates of point Q. We have: P(1, 2, 3) Q(2, 1, 4) For (a): ∆𝑑_a = Q - P = (1, -1, 1) The total distance moved is 2 mm, so we need to normalize the displacement vector and multiply it by 2 mm: ∆𝑑_a = 2 * (1/sqrt(3), -1/sqrt(3), 1/sqrt(3)) = (2/sqrt(3), -2/sqrt(3), 2/sqrt(3)) mm For (b): ∆𝑑_b = P - Q = (-1, 1, -1) The total distance moved is 2 mm, so we need to normalize the displacement vector and multiply it by 2 mm: ∆𝑑_b = 2 * (-1/sqrt(3), 1/sqrt(3), -1/sqrt(3)) = (-2/sqrt(3), 2/sqrt(3), -2/sqrt(3)) mm

Calculate the angle between the electric field vector and the displacement vector

The electric field vector is given by 𝐸 = 120 𝑎_𝜌 V/m. 𝑎_𝜌 is a unit vector pointing in the radial direction, and in Cartesian coordinates, 𝑎_𝜌 = (1, 0, 0). For (a): The angle θ can be found by taking the dot product between 𝐸 and ∆𝑑_a and dividing by the magnitudes of 𝐸 and ∆𝑑_a, then taking the arccosine of the result. cos(θ_a) = (120(1,0,0) • (2/sqrt(3), -2/sqrt(3), 2/sqrt(3))) / (120 * 2) θ_a = arccos(1/sqrt(3)) For (b): Similarly, we find the angle for the displacement vector ∆𝑑_b: cos(θ_b) = (120(1,0,0) • (-2/sqrt(3), 2/sqrt(3), -2/sqrt(3))) / (120 * 2) θ_b = arccos(-1/sqrt(3))

Calculate the incremental work done

To calculate the incremental work done, we use the given formula: ΔW = q * 𝐸 * Δ𝑑 * cos(θ) For (a): ΔW_a = (50 * 10^-6 C) * (120 V/m) * 2 * 0.001 * cos(θ_a) ΔW_a = (50 * 10^-6 C) * (120 V/m) * 0.002 * (1/sqrt(3)) For (b): ΔW_b = (50 * 10^-6 C) * (120 V/m) * 2 * 0.001 * cos(θ_b) ΔW_b = (50 * 10^-6 C) * (120 V/m) * 0.002 * (-1/sqrt(3)) Remember to change the units for your results or write them in scientific notation as necessary.