Problem 19.31 Part A Find the direction and magnitude of the net electrostatic f
ID: 1327440 • Letter: P
Question
Problem 19.31
Part A
Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.3C and d=34cm.
Express your answer using two significant figures.
Part B
Express your answer using three significant figures.
Figure 1 of 1
Problem 19.31
Part A
Find the direction and magnitude of the net electrostatic force exerted on the point charge q2 in the figure.(Figure 1) Let q=+2.3C and d=34cm.
Express your answer using two significant figures.
Fnet = NPart B
Express your answer using three significant figures.
= counterclockwise from q2-q3 directionFigure 1 of 1
Explanation / Answer
here,
Force due to charge q1 and q2
F12 = 2kq/r^2 NC
force due to charge q1 and q4
F14 = 4kq/r^2 Nc
Force due to charge 1 AND 3
F13 = 3kq/r^2
as F13 make 45 degrees ,so writing in its component form
Fx13 = F13 Cos45
Fy13 = F13 Sin45
adding forces according to coordinates
Fx = F14 + Fx13 = 4kq/r^2 + 3*k*q * Cos45 /R13^2
Fy = F12 + Fy13 = 2kq/r^2 + 3*k*q * Sin45 /R13^2
where R13 is sqrt(d^2 + d^2) = sqrt(2*d^2) = sqrt (2) (34 *10^-2) = 0.48 m
Using value of Q = 2.3C and r = d = 34 *10^-2 m
Fx = (4 * 2.3 * 9 * 10^9) / ((34 *10^-2)^2) + ( 3 * 2.3 * 9 * 10^9 * 0.707 ) / (0.48^2)
Fx = 9.06 *10^11 N
Fy = (2 * 9 * 10^9 * 2.3) / ((34 *10^-2)^2) + ( 3 * 9 * 10^9 * 2.3 * 0.707) / (0.48^2)
Fy = 5.48 *10^11 N
Therefore force in x and y direction is 7.38 *10^11 N and 3.8 * 10^11 N respectively
Now,
magnitude of force = sqrt(fx^2 + Fy^2)
Mf = sqrt((9.06 *10^11)^2 + (5.48 *10^11)^2)
Mf = 1.05 * 10^11 N
Therefore magnitude of resultant force is 1.05* 10^11 N
and
tan = y / x = 5.04 * 10^11 / 9.06 *10^11
tan = 4.88 / 9.06 = 0.53
= 27.92 degrees
The angle of resultant force is 29.68 degrees