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Chapter 9: Problem 24

Two negative charges with magnitude \(q\) and \(2 q\) sit at points \((1,0)\) and \((0,1)\) on the \(x\) - and \(y\)-axis, respectively. Which figure best represents the total electric field at the origin? (A) \(\mathrm{A}\) (B) \(\mathrm{B}\) (C) \(\mathrm{C}\) (D) \(\mathrm{D}\) (E) \(\mathrm{E}\)

Short Answer

Expert verified
Question: Describe the direction and magnitude of the total electric field at the origin, considering the presence of a negative charge \(q\) at (1,0) and a negative charge \(2q\) at (0,1). Answer: The total electric field at the origin has a direction between the negative x and y-axes, with a stronger field towards the negative y-axis due to the presence of \(2q\). The magnitude along the x-axis is \(kq\) and along the y-axis is \(2kq\).

Step by step solution

01

Find the electric field due to charge \(q\) at the origin.

First, let's find the electric field at the origin due to the charge \(q\) at point (1,0). The direction of the electric field is always away from a positive charge and towards a negative charge. Since the charge at point (1,0) is negative, the electric field at the origin due to this charge will be directed towards it, i.e., along the negative x-axis. The magnitude of the electric field due to this charge, \(E_q\), can be calculated using Coulomb's law: \(E_q = \frac{kq}{r^2}\) Here, \(k\) is Coulomb's constant, \(q\) is the charge magnitude, and \(r\) is the distance between the charge and the origin. In this case, \(r = 1\), so the electric field magnitude due to this charge is \(E_q = kq.\)
02

Find the electric field due to charge \(2q\) at the origin.

Now let's find the electric field at the origin due to the charge \(2q\) at point (0,1). Since this charge is also negative, the electric field at the origin due to this charge will be directed towards it, i.e., along the negative y-axis. The magnitude of the electric field due to this charge, \(E_{2q}\), can be calculated using Coulomb's law: \(E_{2q} = \frac{k(2q)}{r^2}\) Again, the distance between the charge and the origin is \(r = 1\), so the electric field magnitude due to this charge is \(E_{2q} = 2kq.\)
03

Find the vector sum of the individual electric fields.

We can now find the total electric field at the origin by calculating the vector sum of the electric fields due to each charge. \(E_{total} = \vec{E_q} + \vec{E_{2q}}\) Since the electric field due to the charge at (1,0) is directed along the negative x-axis and the electric field due to the charge at (0,1) is directed along the negative y-axis, we can write the electric field vector components for these charges: \(\vec{E_q} = (-kq, 0)\) \(\vec{E_{2q}} = (0, -2kq)\) \(E_{total} = \vec{E_q} + \vec{E_{2q}} = (-kq, 0) + (0, -2kq) = (-kq, -2kq)\)
04

Describe the total electric field at the origin.

The total electric field at the origin, \(E_{total}\), is given by the vector \((-kq, -2kq)\). This means that the electric field has a magnitude along the x-axis of \(kq\) and a magnitude twice as large along the y-axis (\(2kq\)). In other words, the electric field is pointing in the direction between the negative x and y-axes, with a stronger electric field towards the negative y-axis. The best representation of the total electric field will have a direction between the negative x and y-axes, with a stronger electric field towards the negative y-axis.

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Most popular questions from this chapter

An electron is released from rest just above a very large, negatively charged sheet, which carries surface charge density \(\sigma=5 \mathrm{nC} / \mathrm{m}^2\). When the electron is \(0.5 \mathrm{~m}\) above the sheet its speed is most nearly (A) \(4 \times 10^8 \mathrm{~m} / \mathrm{s}\) (B) \(3 \times 10^8 \mathrm{~m} / \mathrm{s}\) (C) \(5 \times 10^7 \mathrm{~m} / \mathrm{s}\) (D) \(1 \times 10^7 \mathrm{~m} / \mathrm{s}\) (E) \(300 \mathrm{~m} / \mathrm{s}\)

A wire carries a current of \(I=10 \mathrm{~A}\) into the page. With a magnetic field probe, a student measures the \(B\)-field at five points and notes down the results: \((1,2.2) ;(2,0.85) ;(5,0.35) ;(8,0.31) ;(10,0.11)\). The first number in each pair is the distance to the right of the wire in \(\mathrm{cm}\) and the second number is the \(B\)-field in units of \(10^{-4} \mathrm{~T}\). a) Plot the data, then plot the expected curve on the same graph. b) What would you say is the average error of the measurements? c) After you have plotted the data, someone reports that the earth's \(B\)-field at this location is \(0.5\) gauss, pointing to the top of the page. Replot the corrected \(B\)-field. d) What percentage error does the earth's field introduce to the measurements as a function of \(r\) ? e) Would you say the original results were trustworthy?

A proton is launched from a very long negatively charged plate at an initial velocity \(v_o\) and angle \(\theta\) toward an identical, positively charged plate, as shown below. The two plates are held at a potential difference \(V=1000\) volts, and the distance between the two plates is \(d\). a) Draw in the electric field vectors. Write an algebraic expression for the electric field strength in terms of given quantities. b) If \(v_o=4 \times 10^5 \mathrm{~m} / \mathrm{s}\), and \(\theta=60^{\circ}\), does \(--_{-}--_{-}--_{-}--^{-}\) the proton hit the top plate? If so, with what velocity? c) If \(d=25 \mathrm{~cm}\), how long is the proton in flight? d) If a uniform magnetic field pointing along the proton's initial velocity vector is introduced between the plates, does this alter the conclusion to (b)?

A particle of mass \(m\) and charge \(+q\) is shot with velocity \(\mathbf{v}\) into a region of uniform magnetic field \(\mathbf{B}\). Suppose that \(\mathbf{v}\) points to the top of the page and \(\mathbf{B}\) points out of the page. Ignoring gravity, the particle (A) travels in a straight line. (B) moves clockwise in a circle with radius \(r=q B / m v\). (C) moves counterclockwise in a circle with radius \(r=q B / m v\). (D) moves clockwise in a circle with radius \(r=m v / q B\). (E) moves counterclockwise in a circle with radius \(r=m v / q B\).

The positively charged rod is brought near the large sphere, but without touching it. The two spheres are separated and lastly the rod removed to a distance. We can then say that (A) the big sphere will be positively charged and the smaller sphere will be negatively charged. (B) both spheres will be positively charged. (C) the big sphere will be negatively charged and the smaller sphere will be positively charged. (D) both spheres will be negatively charged. (E) all the charge will migrate to the smaller sphere.
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