In the figure below, each charged particle is located at one of the four vertices of a square with side length = a. In the figure, A = 4, B = 5, and C = 7, and q > 0. ☐ Bq 9 (a) What is the expression for the magnitude of the electric field in the upper right corner of the square (at the location of q)? (Use the following as necessary: q, a, and k.) E = ke V(2√29-59)²+(2√29-79)2 a² Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric field at this location. 39.19 x (counterclockwise from the +x-axis) (b) Determine the expression for the total electric force exerted on the charge q. (Enter the magnitude. Use the following as necessary: q, a, and k₂-) F=gk √(2√29-59)+(2√29-79) 2 a² Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric force on q. 39.9 x (counterclockwise from the +x-axis) (c) What If? How would the answers to parts (a) and (b) change if each of the four charges were negative with the same magnitude? Select all that apply. The electric field would be the same magnitude and direction as the field in part (a). The force would be the same magnitude and direction as the force in part (b). The force would be the same magnitude but opposite direction as the force in part (b). The electric field would be the same magnitude but opposite direction as the field in part (a).
In the figure below, each charged particle is located at one of the four vertices of a square with side length = a. In the figure, A = 4, B = 5, and C = 7, and q > 0. ☐ Bq 9 (a) What is the expression for the magnitude of the electric field in the upper right corner of the square (at the location of q)? (Use the following as necessary: q, a, and k.) E = ke V(2√29-59)²+(2√29-79)2 a² Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric field at this location. 39.19 x (counterclockwise from the +x-axis) (b) Determine the expression for the total electric force exerted on the charge q. (Enter the magnitude. Use the following as necessary: q, a, and k₂-) F=gk √(2√29-59)+(2√29-79) 2 a² Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric force on q. 39.9 x (counterclockwise from the +x-axis) (c) What If? How would the answers to parts (a) and (b) change if each of the four charges were negative with the same magnitude? Select all that apply. The electric field would be the same magnitude and direction as the field in part (a). The force would be the same magnitude and direction as the force in part (b). The force would be the same magnitude but opposite direction as the force in part (b). The electric field would be the same magnitude but opposite direction as the field in part (a).
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Transcribed Image Text:In the figure below, each charged particle is located at one of the four vertices of a square with side length = a. In the figure, A = 4, B = 5, and C = 7, and q > 0.
☐
Bq
9
(a) What is the expression for the magnitude of the electric field in the upper right corner of the square (at the location of q)? (Use the following as necessary: q, a, and k.)
E = ke
V(2√29-59)²+(2√29-79)2
a²
Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric field at this location.
39.19
x (counterclockwise from the +x-axis)
(b) Determine the expression for the total electric force exerted on the charge q. (Enter the magnitude. Use the following as necessary: q, a, and k₂-)
F=gk
√(2√29-59)+(2√29-79) 2
a²
Give the direction angle (in degrees counterclockwise from the +x-axis) of the electric force on q.
39.9
x
(counterclockwise from the +x-axis)
(c) What If? How would the answers to parts (a) and (b) change if each of the four charges were negative with the same magnitude? Select all that apply.
The electric field would be the same magnitude and direction as the field in part (a).
The force would be the same magnitude and direction as the force in part (b).
The force would be the same magnitude but opposite direction as the force in part (b).
The electric field would be the same magnitude but opposite direction as the field in part (a).
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