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Chapter 4: Problem 3

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

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

Step by step solution

01

Calculate the displacement vector

To calculate the displacement vector 鈭嗮潙� 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): 鈭嗮潙慱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: 鈭嗮潙慱a = 2 * (1/sqrt(3), -1/sqrt(3), 1/sqrt(3)) = (2/sqrt(3), -2/sqrt(3), 2/sqrt(3)) mm For (b): 鈭嗮潙慱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: 鈭嗮潙慱b = 2 * (-1/sqrt(3), 1/sqrt(3), -1/sqrt(3)) = (-2/sqrt(3), 2/sqrt(3), -2/sqrt(3)) mm
02

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

The electric field vector is given by 饾惛 = 120 饾憥_饾湆 V/m. 饾憥_饾湆 is a unit vector pointing in the radial direction, and in Cartesian coordinates, 饾憥_饾湆 = (1, 0, 0). For (a): The angle 胃 can be found by taking the dot product between 饾惛 and 鈭嗮潙慱a and dividing by the magnitudes of 饾惛 and 鈭嗮潙慱a, then taking the arccosine of the result. cos(胃_a) = (120(1,0,0) 鈥� (2/sqrt(3), -2/sqrt(3), 2/sqrt(3))) / (120 * 2) 胃_a = arccos(1/sqrt(3)) For (b): Similarly, we find the angle for the displacement vector 鈭嗮潙慱b: cos(胃_b) = (120(1,0,0) 鈥� (-2/sqrt(3), 2/sqrt(3), -2/sqrt(3))) / (120 * 2) 胃_b = arccos(-1/sqrt(3))
03

Calculate the incremental work done

To calculate the incremental work done, we use the given formula: 螖W = q * 饾惛 * 螖饾憫 * cos(胃) For (a): 螖W_a = (50 * 10^-6 C) * (120 V/m) * 2 * 0.001 * cos(胃_a) 螖W_a = (50 * 10^-6 C) * (120 V/m) * 0.002 * (1/sqrt(3)) For (b): 螖W_b = (50 * 10^-6 C) * (120 V/m) * 2 * 0.001 * cos(胃_b) 螖W_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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 蟻. 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.

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Most popular questions from this chapter

Two point charges, \(1 \mathrm{nC}\) at \((0,0,0.1)\) and \(-1 \mathrm{nC}\) at \((0,0,-0.1)\), are in free space. \((a)\) Calculate \(V\) at \(P(0.3,0,0.4) .\) (b) Calculate \(|\mathbf{E}|\) at \(P .(c)\) Now treat the two charges as a dipole at the origin and find \(V\) at \(P\).

A sphere of radius \(a\) carries a surface charge density of \(\rho_{s 0} \mathrm{C} / \mathrm{m}^{2} .(a)\) Find the absolute potential at the sphere surface. ( \(b\) ) A grounded conducting shell of radius \(b\) where \(b>a\) is now positioned around the charged sphere. What is the potential at the inner sphere surface in this case?

A certain spherically symmetric charge configuration in free space produces an electric field given in spherical coordinates by $$ \mathbf{E}(r)=\left\\{\begin{array}{ll} \left(\rho_{0} r^{2}\right) /\left(100 \epsilon_{0}\right) \mathbf{a}_{r} \mathrm{~V} / \mathrm{m} & (r \leq 10) \\ \left(100 \rho_{0}\right) /\left(\epsilon_{0} r^{2}\right) \mathrm{a}_{r} \mathrm{~V} / \mathrm{m} & (r \geq 10) \end{array}\right. $$ where \(\rho_{0}\) is a constant. (a) Find the charge density as a function of position. (b) Find the absolute potential as a function of position in the two regions, \(r \leq 10\) and \(r \geq 10 .(c)\) Check your result of part \(b\) by using the gradient. (d) Find the stored energy in the charge by an integral of the form of Eq. (43). (e) Find the stored energy in the field by an integral of the form of Eq. (45).

Let a uniform surface charge density of \(5 \mathrm{nC} / \mathrm{m}^{2}\) be present at the \(z=0\) plane, a uniform line charge density of \(8 \mathrm{nC} / \mathrm{m}\) be located at \(x=0, z=4\), and a point charge of \(2 \mu \mathrm{C}\) be present at \(P(2,0,0) .\) If \(V=0\) at \(M(0,0,5)\), find \(V\) at \(N(1,2,3)\).

Compute the value of \(\int_{A}^{P} \mathbf{G} \cdot d \mathbf{L}\) for \(\mathbf{G}=2 y \mathbf{a}_{x}\) with \(A(1,-1,2)\) and \(P(2,1,2)\) using the path ( \(a\) ) straight-line segments \(A(1,-1,2)\) to \(B(1,1,2)\) to \(P(2,1,2) ;(b)\) straight-line segments \(A(1,-1,2)\) to \(C(2,-1,2)\) to \(P(2,1,2)\)
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