Charges q_1 = -2.90 mu C and q_2 = +5.50 mu C are located as shown in the figure
ID: 1572629 • Letter: C
Question
Charges q_1 = -2.90 mu C and q_2 = +5.50 mu C are located as shown in the figure, where d = 0.505 m. Point A is located at (4/5 d, 2/3 d) Determine the magnitude of the electric field at point A. Two electric force vectors act on a particle. Their x-components are 14.5 N and -7.65 N and their/-components are -14.0 N and -4.70 N, respectively. For the resultant electric force, find the following. the x-component the y-component the magnitude of the resultant electric force the direction of the resultant electric force, measured counterclockwise from the positive x-axis counterclockwise from the +x-axis Two protons are separated by a distance of 0.266 m. Given the proton charge of 1.60 times 10^-19 C, determine the magnitude of the electric force that one proton exerts on the other.Explanation / Answer
0.003)
The net x component is
(a)
Fx= 14.5 N-7.65 N = 6.85 N i
(b)
Fy =-14 N-4.7 N = -18.7 N j
(c)
magnitude of force is
F = sqrt Fx^2 + Fy^2
= sqrt (6.85)^2 + (-18.7)^2
=19.91 N
(d)
direction
theta = tan^-1 ( -18.7/6.85) =290.11 degree counter clockwise from + x axis
0.008)
F = kq^2/r^2
= 9 * 10^9 ( 1.6 * 10^-19)^2/(0.265)^2
=3.28 * 10^-27 N
As per guide lines I worked two problem kindly please post remaining posts in the next post