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Chapter 2: Problem 3

Point charges of \(50 \mathrm{nC}\) each are located at \(A(1,0,0), B(-1,0,0), C(0,1,0)\), and \(D(0,-1,0)\) in free space. Find the total force on the charge at \(A\).

Short Answer

Expert verified
Answer: The total force acting on the point charge at A is 5.62 x 10^-8 N, and it acts along the x-axis.

Step by step solution

01

Calculate the distance between charges

We will find the distance between point A and other points B, C, and D using the distance formula for 3D coordinates: \(r = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\) Distance between A and B, \(r_{AB} = \sqrt{((-1-1)^2 + (0-0)^2 + (0-0)^2)} = 2\) Distance between A and C, \(r_{AC} = \sqrt{((0-1)^2 + (1-0)^2 + (0-0)^2)} = \sqrt{2}\) Distance between A and D, \(r_{AD} = \sqrt{((0-1)^2 + (-1-0)^2 + (0-0)^2)} = \sqrt{2}\)
02

Calculate individual forces acting on A

Now using Coulomb's law formula, we will calculate the forces. Force due to B on A, \(F_{AB} = k \frac{q_A q_B}{r_{AB}^2}\) \(F_{AB} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(2)^2} = 5.62 \times 10^{-8} N\) Force due to C on A, \(F_{AC} = k \frac{q_A q_C}{r_{AC}^2}\) \(F_{AC} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(\sqrt{2})^2} = 1.124 \times 10^{-7} N\) Force due to D on A, \(F_{AD} = k \frac{q_A q_D}{r_{AD}^2}\) \(F_{AD} = (8.99 \times 10^9) \frac{(50 \times 10^{-9})(50 \times 10^{-9})}{(\sqrt{2})^2} = 1.124 \times 10^{-7} N\)
03

Calculate the total force on A

To find the total force on A, we need to add the forces acting on A due to B, C, and D. Since the forces due to C and D are equal and along the y-axis, their combined effect along the y-axis will cancel each other out. Thus, the total force on A will be along the x-axis and equal to the force exerted by B. Total Force on A = \(F_{AB}\) Total Force on A = \(5.62 \times 10^{-8} N\) So, the total force on charge A is \(5.62 \times 10^{-8} N\) along the x-axis.

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Most popular questions from this chapter

(a) Find the electric field on the \(z\) axis produced by an annular ring of uniform surface charge density \(\rho_{s}\) in free space. The ring occupies the region \(z=0, a \leq \rho \leq b, 0 \leq \phi \leq 2 \pi\) in cylindrical coordinates. \((b)\) From your part (a) result, obtain the field of an infinite uniform sheet charge by taking appropriate limits.

The electron beam in a certain cathode ray tube possesses cylindrical symmetry, and the charge density is represented by \(\rho_{v}=-0.1 /\left(\rho^{2}+10^{-8}\right)\) \(\mathrm{pC} / \mathrm{m}^{3}\) for \(0<\rho<3 \times 10^{-4} \mathrm{~m}\), and \(\rho_{v}=0\) for \(\rho>3 \times 10^{-4} \mathrm{~m} .(a)\) Find the total charge per meter along the length of the beam; \((b)\) if the electron velocity is \(5 \times 10^{7} \mathrm{~m} / \mathrm{s}\), and with one ampere defined as \(1 \mathrm{C} / \mathrm{s}\), find the beam current.

A \(100-n C\) point charge is located at \(A(-1,1,3)\) in free space. \((a)\) Find the locus of all points \(P(x, y, z)\) at which \(E_{x}=500 \mathrm{~V} / \mathrm{m} \cdot(b)\) Find \(y_{1}\) if \(P\left(-2, y_{1}, 3\right)\) lies on that locus.

Two identical uniform sheet charges with \(\rho_{s}=100 \mathrm{nC} / \mathrm{m}^{2}\) are located in free space at \(z=\pm 2.0 \mathrm{~cm}\). What force per unit area does each sheet exert on the other?

A uniform line charge of \(2 \mu \mathrm{C} / \mathrm{m}\) is located on the \(z\) axis. Find \(\mathbf{E}\) in rectangular coordinates at \(P(1,2,3)\) if the charge exists from \((a)-\infty<\) \(z<\infty ;(b)-4 \leq z \leq 4\).
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