College Physics
11th Edition
ISBN: 9781305952300
Author: Raymond A. Serway, Chris Vuille
Publisher: Cengage Learning
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- Two point charges, Q₁ = -5.0 μC and Q2 = 2.0 μC are located between two oppositely charged parallel plates, as shown in the figure(Figure 1). The two charges are separated by a distance of x = 0.32 m. Assume that the electric field produced by the charged plates is uniform and equal to E = 7.4x104 N/C. Part A Calculate the magnitude of the net electrostatic force on Q1. Express your answer using two significant figures. IVE ΑΣΦ F = Submit Part B Request Answer Give its direction. O to the left O to the right Submit Request Answer ? Narrow_forwardPart A An electric dipole is formed from two charges, +q, spaced 0.600 cm apart. The dipole is at the origin, oriented along the y-axis. The electric field strength at the point (r, y) = (0 cm, 10 cm) is 340 N/C What is the charge q? Give your answer in nC. Express your answer in nanocoulombs to three significant figures. > View Available Hint(s) ΑΣφ q =160 nC Submit Previous Answers Incorrect; Try Again; 3 attempts remaining Part B What is the electric field strength at the point (x, y) = (10 cm, 0 cm)? Express your answer with the appropriate units. View Available Hint(s) HA E= Value Units Submitarrow_forwardPart A An electron moving parallel to a uniform electric field increases its speed from 2.0x10' m/s to 4.0×10 m/s over a distance of 1.2 cm. What is the electric field strength? Express your answer with the appropriate units. HA E = Value Units Submit Request Answerarrow_forward
- Part A and Barrow_forwardElectric Fields 1. A wire has been given a net positive electric charge Q evenly distributed along its length. The wire has a length of L and is located on the x-axis with its left end at the origin as drawn. Set up the integral you need to do to find the electric field at a distance d to the left of the origin. Some steps you may wish to follow are: a. Figure out the direction the electric field will have at our location x = -d. This will be the same as f b. Write the differential charge dą in terms of the charge per length and a differential length. c. ris a function of x which means we have to integrate to find the electric field. Write down r. d. Put everything together (including the vector direction) into one integral that would allow you to find the E-field. Write this in the box below. e. You are not required to complete the integral and no credit is given for completing the integral. d dE =arrow_forwardneed help on part A, C, Darrow_forward
- The electric field 0.340 m from a very long uniform line of charge is 760 N/C. Part A How much charge is contained in a section of the line of length 2.80 cm ? lgl = C Submit Request Answer Provide Feedbackarrow_forwardProblem 1 Shown on the right is an electroscope. The sphere on the top, the leaves at the bot- tom, and the rod connecting them are made by conducting material. The spherical shell that blocks the influence of external electric fields is also made by conducting material but the contact with the rod is insulated, so that charges will not shuffle between the rod and the spherical shell. The electroscope is first negatively charged by touching the sphere with a negatively charged plastic plate. The leaves spread apart and we remove the plastic plate. We then bring a positively charged glass bar close to the sphere without touching the sphere. What happens to the leaves? Leaves (A) One leaf moves higher, the other lower. (B) Nothing changes to the leaves (C) The leaves spread farther apart. (D) The leaves get closer together. Clearly specify/circle your answer, and give a brief explanation.arrow_forward
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