Question

  

12. In the figure below, d = 2.8 mm. What are the magnitude and direction (angle) of the electric force on the +5 nC charge?

10nC ------ d ------ -5nC

I

2d I

I

+5 nC

Answer 1

q1 = 10 nC

q2 = -5 nC

q3 = + 5nC

q1 exerts a force of repulsion on q3

force exerted by q1 on q3 F13y = -k*q1*q3/r13^2

r13 = 2d = 5.6 mm = 5.6*10^-3 m

F13y = -(9*10^9*10*10^-9*5*10^-9)/(5.6*10^-3)^2 = -0.0143 N

F13x = 0

tantheta = 2d/d = 2

theta = 63.43 degrees

r13 = sqrt(d^2+(2d)^2 ) = d*sqrt5 = 6.26 mm

q2 exerts a force of attraction on q3

force exerted by q2 on q3  

F23x = k*q1*q3*costheta/r13^2 = 9*10^9*5*10^-9*5*10^-9*cos63.43/(6.26*10^-3)^2 = 0.00267 N

F23y = k*q1*q3*sintheta/r13^2 = 9*10^9*5*10^-9*5*10^-9*sin63.43/(6.26*10^-3)^2 = 0.00513 N

Fx = F13x + F23x = 0.00267 N

Fy = F13y + F23y = -0.00917 N

magnitude = sqrt(0.00267^2+0.00917^2) = 0.00955 N

direction = tan^-1(Fy/Fx) = 73.8 degrees