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Chapter 22: Problem 25

\( \mathrm{~A}+48.00-\mathrm{nC}\) point charge is placed on the \(x\) -axis at \(x=4.000 \mathrm{~m},\) and \(\mathrm{a}-24.00-\mathrm{n} \mathrm{C}\) point charge is placed on the \(y\) -axis at \(y=-6.000 \mathrm{~m} .\) What is the direction of the electric field at the origin?

Short Answer

Expert verified
Answer: The direction of the net electric field at the origin can be found using the formula \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\), where \(E_A\) and \(E_B\) are the electric fields due to charges A and B, respectively. Plug in the values of A and B, calculate \(E_A\) and \(E_B\), and then find the angle \(\theta\) to determine the direction of the net electric field at the origin.

Step by step solution

01

Finding the electric field due to charge A at the origin

To find the electric field due to the first charge A at the origin, use the formula for the electric field from a point charge: \(E = k* \frac{Q}{r^2}\), where \(k = 8.9875517923*10^9 N m^2/C^2\) is Coulomb's constant, \(Q\) is the charge, and \(r\) is the distance between the charge and the point at which we want to find the electric field. The distance between charge A and the origin is \(x=4m\). Therefore, the electric field at the origin due to charge A is: \(E_A = k* \frac{A}{(4m)^2}\).
02

Finding the electric field due to charge B at the origin

Repeat the process for charge B, which is placed on the y-axis at \(y=-6m\). The distance between charge B and the origin is \(6m\). Therefore, the electric field at the origin due to charge B is: \(E_B = k* \frac{B}{(6m)^2}\), where B is -24.00 nC.
03

Determining the components of the electric fields

We know that the electric field due to charge A is in the negative x-direction and the electric field due to charge B is in the positive y-direction. Thus, the components of the electric fields are: \(E_{Ax}=-E_A, E_{Ay}=0\) \(E_{Bx}=0, E_{By}=E_B\)
04

Vector addition of the electric fields

Now, we'll find the net electric field at the origin by adding the components of the electric fields due to charges A and B: \(E_{netx} = E_{Ax} + E_{Bx} = -E_A\) \(E_{nety} = E_{Ay} + E_{By} = E_B\)
05

Calculating the direction of the net electric field

To find the direction of the net electric field at the origin, we need to determine the angle with respect to the x-axis. We use the arctangent function to find the angle: \(\theta = \arctan(\frac{E_{nety}}{E_{netx}}) = \arctan(\frac{E_B}{-E_A})\) Now, you can plug in the values of A, B and find \(E_A\) and \(E_B\) to calculate the angle, \(\theta\), which gives the direction of the net electric field at the origin.

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Most popular questions from this chapter

A solid, nonconducting sphere of radius \(a\) has total charge \(Q\) and a uniform charge distribution. Using Gauss's Law, determine the electric field (as a vector) in the regions \(ra\) in terms of \(Q\).

Consider a long horizontally oriented conducting wire with \(\lambda=4.81 \cdot 10^{-12} \mathrm{C} / \mathrm{m} .\) A proton \(\left(\mathrm{mass}=1.67 \cdot 10^{-27} \mathrm{~kg}\right)\) is placed \(0.620 \mathrm{~m}\) above the wire and released. What is the magnitude of the initial acceleration of the proton?

To be able to calculate the electric field created by a known distribution of charge using Gauss's Law, which of the following must be true? a) The charge distribution must be in a nonconducting medium. b) The charge distribution must be in a conducting medium. c) The charge distribution must have spherical or cylindrical symmetry. d) The charge distribution must be uniform. e) The charge distribution must have a high degree of symmetry that allows assumptions about the symmetry of its electric field to be made.

A proton enters the gap between a pair of metal plates (an electrostatic separator) that produces a uniform, vertical electric field between them. Ignore the effect of gravity on the proton. a) Assuming that the length of the plates is \(15.0 \mathrm{~cm}\), and that the proton will approach the plates at a speed of \(15.0 \mathrm{~km} / \mathrm{s}\) what electric field strength should the plates be designed to provide, if the proton must be deflected vertically by \(1.50 \cdot 10^{-3} \mathrm{rad} ?\) b) What speed does the proton have after exiting the electric field? c) Suppose the proton is one in a beam of protons that has been contaminated with positively charged kaons, particles whose mass is \(494 \mathrm{MeV} / \mathrm{c}^{2}\left(8.81 \cdot 10^{-28} \mathrm{~kg}\right)\), compared to the mass of the proton, which is \(938 \mathrm{MeV} / \mathrm{c}^{2}\left(1.67 \cdot 10^{-27} \mathrm{~kg}\right)\) The kaons have \(+1 e\) charge, just like the protons. If the electrostatic separator is designed to give the protons a deflection of \(1.20 \cdot 10^{-3} \mathrm{rad}\), what deflection will kaons with the same momentum as the protons experience?

A single positive point charge, \(q,\) is at one corner of a cube with sides of length \(L\), as shown in the figure. The net electric flux through the three net electric flux through the three adjacent sides is zero. The net electric flux through each of the other three sides is a) \(q / 3 \epsilon_{0}\). b) \(q / 6 \epsilon_{0}\). c) \(q / 24 \epsilon_{0}\). d) \(q / 8 \epsilon_{0}\).
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