1 st Midterm Exam11- chapter23-question22Two point charges are separated by 25.0 cm (see Figure).Find the magnitude of the net electric field (in N/C) these-6.25 nC- 12.5 nCcharges produce at point A (k 8.99 x 10° N m2/C2)10.0 cm10.0 cm25.0 cmOption1Option2Option3Option4Option52.74 x 10 4.74 x 103 6.74 x 103 8.74 x 103 10.74 x 10a)Option2b)Option3Option1Option5e)Option4Leave blankAnswer1.

Given: Q1 = -6.25 nC; Q2 = -125nC; And k= 8.99 x 109 N .m2/C2

11 - chapter23-question22 Two point charges are separated by 25.0 cm (see Figure). Find the magnitude of the net electric field (in N/C) these charges produce at point A (k = 8.99 x 10° N m² /C²) -6.25 nC -12.5 nC 10.0 cm 10.0 cm 25.0 cm Option1 Option2 Option3 Option4 Option5 2.74 x 10 4.74 x 10 6.74 x 10 8.74 x 103 10.74 x 10 a) O Option2 b) Option3 c) Option1 d) Option5 e) Option4 Leave blank Ans

11 - chapter23-question22 Two point charges are separated by 25.0 cm (see Figure). Find the magnitude of the net electric field (in N/C) these charges produce at point A (k = 8.99 x 10° N m² /C²) -6.25 nC -12.5 nC 10.0 cm 10.0 cm 25.0 cm Option1 Option2 Option3 Option4 Option5 2.74 x 10 4.74 x 10 6.74 x 10 8.74 x 103 10.74 x 10 a) O Option2 b) Option3 c) Option1 d) Option5 e) Option4 Leave blank Ans

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)...

See similar textbooks

Similar questions

1.
1. A point charge ql = -6.0 µC is at the point x = - 0.800 m, y = 0.600 m and a second point charge q2 = - 5.0 µC is at the post x = - 0.800 m, y = 0 m. Calculate the magnitude and direction of the net electric force exerted by ql and q2 on the third charge, q3 = - 2.0 nC located at the origin. 2. In problem #1, find the total electric field at the origin.
At what distance from a particle with a charge of 5.00 nC does the electric field of that charge have a magnitude of 3 N/C? The distance = Answer m
A point charge gives rise to an electric field with magnitude 4 N/C at a distance of 8 m. If the distance is increased to 16 m, then what will be the new magnitude of the electric field? N/C
ZA 1.5 m 2.0 m 1.0 m The electric field in a region is given by E = a/(h+ cx)î where a= 200 , b=1.0 , and c=2.0 What is the net charge enclosed by the shaded volume shown above? Your answer: O A) -8,8 x 10-10 C O B) -70,5 x 10-10 c O C) 0 O D) -21,2 x10-10 c O E) 15 x10-10 C
Let q 1 = 2 μC, q2 = -4 μC, q3 = = +3 μC, q4 = -5 μC, q5 = +9 μC.a. What is the net force on q 5 ?b. What is the electric field at the location of q 5 due to the other 4 charges?c. What would be the force on q 5 if you changed it to 7 μC?
SET 1 1. Determine the electric force on the -5 µC charge due to the other two ch as shown in Figure 1. - 5 µC 40° 7 cm Figure 1 6 cm - 3 µC +6 μC
What is the magnitude of the electric field due to a 6.0 x 10-⁹ C charge at a point located 0.025 m from the charge? A) 8.6 x 104 N/C B) 1.2 x 10³ N/C C) 1.8 x 105 N/C D) 3.6 x 106 N/C
A point charge of -5µC is located at x = 4 m, y = -2 m. A second point charge of 12µC is located at æ =1 m, y = 2 m. What is the magnitude of the electric field at x = -1 m, y = 0? А. — 12.9 N/C B. –12.9 kN/C С. 12.9 N/C D. 12.9 kN/C
If the electric field created by an insulated charged conductive sphere with a radius of 8 cm at a distance of 12 cm from its center is 3x104 N/ C, what is the charge density (in uC/ m? unit) on the unit surface of this sphere? A) 1.5 B) 0.6 C) 1.2 D) 1.7 E) 0.3
A point charge gives rise to an electric field with magnitude 4 N/C at a distance of 9 m. If the distance is increased to 45 m, then what will be the new magnitude of the electric field? N/C
a K 6.0 cm 15.0 cm 1 a 10.0 cm 15.0 cm b 5.0 cm 42 Two point charges q1 and q2 of -300 nC and 300 nC, respectively, are separated by a distance of 0.15 m. Calculate the electric field caused by q1, the field caused by q2, and the total field. a) at point a Eq₁ = b) at point c Eq₁ = Eq₂=_ , Eq₂=. _ET=_ _E₁=