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.

Key concepts

Displacement Vector

In physics, a displacement vector is crucial as it helps us to understand the movement from one point to another in space. It is defined as the difference between the final and initial position vectors. To get the displacement vector between two points, we subtract the initial position's coordinates from the final position's coordinates.
In the given exercise, for paths (a) and (b), we have:

• Point P: (1, 2, 3)
• Point Q: (2, 1, 4)

Therefore, the displacement vectors were calculated as:

• From P to Q: \( \Delta \mathbf{d}_a = Q - P = (1, -1, 1) \)
• Normalized and scaled by 2 mm: \( \Delta \mathbf{d}_a = \frac{2}{\sqrt{3}}(1, -1, 1) \)
• From Q to P: \( \Delta \mathbf{d}_b = P - Q = (-1, 1, -1) \)
• Normalized and scaled by 2 mm: \( \Delta \mathbf{d}_b = -\frac{2}{\sqrt{3}}(1, -1, 1) \)

This step-by-step subtraction and normalization help clarify how movement occurs within an electromagnetic field.

Electric Field

The electric field is a vector field around a charged particle or object, representing the field of force exerted on other charges. It is measured in volts per meter (V/m) and dictates how other charges move when placed within the field.
In this problem, the electric field is given as: \( \mathbf{E} = 120 \mathbf{a}_{\rho} \ \text{V/m} \) . Here, \( \mathbf{a}_{\rho} \) is a radial unit vector, aligned in the direction of increasing radial coordinate ρ. It's important to consider the orientation of this field when calculating work done.
This field acts on any charge located within it, producing a force that influences the displacement of the charge.

Work Done on Charge

Work done on a charge in the presence of an electric field can be understood through the equation \(\Delta W = q * E * \Delta d * \cos(\theta)\). Here, \( q \) is the charge, \( E \) is the magnitude of the electric field, \( \Delta d \) is the displacement, and \( \theta \) is the angle between the field and the displacement vector.

• For path (a), the work is calculated as: \( \Delta W_a = (50 \times 10^{-6} \ \text{C}) \times (120 \ \text{V/m}) \times 0.002 \ \times \frac{1}{\sqrt{3}} \)
• For path (b), it's calculated as: \( \Delta W_b = (50 \times 10^{-6} \ \text{C}) \times (120 \ \text{V/m}) \times 0.002 \ \times -\frac{1}{\sqrt{3}} \)

It's crucial to recognize how orientation affects the work performed, specifically through the cosine component representing the directional relationship.

Vector Calculus

Vector calculus is a mathematical tool essential for studying fields and their interactions. It involves operations like dot product, which is used to find the angle between vectors, such as displacement and electric field vectors in this instance.
The dot product, given by \( \mathbf{A} \cdot \mathbf{B} = |\mathbf{A}| |\mathbf{B}| \cos(\theta) \) , allows us to determine the alignment or angle between two vectors. This aspect was used to find the angle \( \theta \) for the displacement vectors in the exercise:

• For path (a), \( \cos(\theta_a) = \frac{120(1,0,0) \cdot (2/\sqrt{3}, -2/\sqrt{3}, 2/\sqrt{3})}{120 \times 2} \)
• For path (b), \( \cos(\theta_b) = \frac{120(1,0,0) \cdot (-2/\sqrt{3}, 2/\sqrt{3}, -2/\sqrt{3})}{120 \times 2} \)

Understanding vector calculus allows one to predict and measure interactions in electromagnetic fields analytically.