Part A Find the direction and magnitude of the net electrostatic force exerted o
ID: 2125174 • Letter: P
Question
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=+2.2?C and d=31cm.
Express your answer using two significant figures.
Fnet= ?N
Part B
Express your answer using three significant figures.
? =
? = ? ? counterclockwise from q2-q3 direction
Explanation / Answer
The force on 2 by1 F21 = k*q1*q2/r^2
= 9.0x10^9*(2.2x10^-6)*(4.4x10^-6)/(0.31^2) = 0.907N +y direction
F23 = k*q2*q3/r^2
= 9.0x10^9*(4.4x10^-6)*(6.6x10^-6)/(0.31^2) = 2.72N +x direction
The distance r24 = 0.31*sqrt(2) = 0.438m
So F24 = k*q2*q4/r^2
= 9.0x10^9*(4.4x10^-6)*(8.8x10^-6)/(0.438^) = 1.82N directed down and to the rt at 45o
Now adding components Fx = 2.72 + 1.82*cos(45) = 4.01N
Fy = 0.907+1.82*sin(45) = 2.19N
So magnitude = sqrt(4.01^2 + 2.19^2) = 4.569N
and ? = arctan(2.19/4.01) = 28.641 degrees