In the figure below, the particles have charges q_1 = -q_2 = 255 nC and q_3 = -q
ID: 1636756 • Letter: I
Question
In the figure below, the particles have charges q_1 = -q_2 = 255 nC and q_3 = -q_4 = 168 nC, and distance a = 3.6 cm. For the following questions, take right to be the +x direction and up to be the +y direction. (a) What is the x N component of the net electrostatic force on particle 3? (b) what is the y N component of the net electrostatic force on particle 3? Did you draw the 3 particles on an x axis? Did you draw the force vectors on the third particle, with the tail of the orientations correct to give a net force of zero? If the third particle is positive, what are the signs of the other 2 the 2 force magnitudes? Did you set up an equation expressing that relation? Did you plug in correctly for the distanceExplanation / Answer
q1=-q2=255nC ; q3=-q4=168nC
Charge q3 will experience force in x direction due to charges q2 and q4
Force on q3 due to q2 = (9*109*255*10-9*168*10-9)/(2*0.0362) = 0.14875N (making angle 45o anticlockwise from positive x axis)
x component of this force = 0.14875cos45o = 0.105 N
Force on q3 due to q4 = (9*109*168*10-9*168*10-9)/(0.0362) =0.196 N(towards postive x direction)
Total x component of forces = 0.105+0.196 =0.301 N(towards positive x direction)
Charge q3 will experience force in y direction due to charges q1 and q2
Force on q3 due to q1 = (9*109*255*10-9*168*10-9)/(0.0362) = 0.2975 N(towards negative y direction)
Force on q3 due to q2 = (9*109*255*10-9*168*10-9)/(2*0.0362) = 0.14875N (making angle 45o anticlockwise from positive x axis)
y component of this force = 0.14875sin45o = 0.105 N(towards positive y direction)
Total y component of forces =-0.2975+0.105 =-0.1925 N=0.1925N(towards negative y direction)