Three charged particles form a triangle particle 1 with char

Three charged particles form a triangle: particle 1 with charge Q_1 = 81.0 nC is at xy coordinates (0, 3.70 mm), particle 2 with charge Q_2 is at (0, -3.70 mm), and particle 3 with charge q = 19.0 nC is at (5.90 mm, 0). What are (a) the x component and (b) the y component of electrostatic force on particle 3 due to the other two particles if Q_1 is equal to 81.0 nC? What are (c) the x component and (d) the y component of electrostatic force on particle 3 due to the other two particles if Q_2 is equal to -81.0 nC, ?

Solution

a) Force on Q3 due to Q1:

both are +ve so force will be repulsive.

and direction of force is along the line joining the charges.

angle with x axis by force = - tan^-1(3.70/5.90) = -32 deg

F31 = kq1q3 / d^2    (cos@ i + sin@j)

F31 =(9 x 10^9 x 81 x 10^-9 x 19 x 10^-9 / (0.0037^2 + 0.0059^2)) (cos(-32)i + sin(-32)j)

F31 = 0.2421i - 0.1513 j N

force on q3 due to q2:

@ = 32 deg

F32 = (9 x 10^9 x 81 x 10^-9 x 19 x 10^-9 / (0.0037^2 + 0.0059^2)) (cos(32)i + sin(32)j)

F32 = + 0.2421i + 0.1513 j N

Fnet = F31 + F32 = 0.484i + 0j

a) x - compo = 0.484 N

b) y - compo = 0

c) angle with x axis by force = - tan^-1(3.70/5.90) = -32 deg

F31 = kq1q3 / d^2    (cos@ i + sin@j)

F31 =(9 x 10^9 x 81 x 10^-9 x 19 x 10^-9 / (0.0037^2 + 0.0059^2)) (cos(-32)i + sin(-32)j)

F31 = 0.2421i - 0.1513 j N

force on q3 due to q2:

@ = 180 + 32 deg = 212 deg

F32 = (9 x 10^9 x 81 x 10^-9 x 19 x 10^-9 / (0.0037^2 + 0.0059^2)) (cos(212)i + sin212)j)

F32 = - 0.2421i - 0.1513 j N

Fnet = F31 + F32 = 0i - 0.3026j N

x-comp= 0 N

y -compo = - 0.3026 N