Three charges are placed as shown The value of q1 is 15 C an

Three charges are placed as shown. The value of q1 is 1.5 ?C and the value of q3 is -3.0 ?C. The net force on q3 is in the negative x-direction.

A)Determine the charge on q2.
Unit hint: use u for ?

B)Determine the magnitude of the net force on q3.

force on q3 due to q1:

magnitude = F31 = k q1 q3 / d^2

= (9 x 10^9 x 1.5 x 10^-6 x 3 x 10^-6) / (4.1^2)

= 0.00241 N

direction - > from q3 to q1.

angle with +ve x axis = 180 + @1

and cos@1 = (5.2^2 + 4.1^2 - 3.3^2) / ( 2 x 4.1 x5.2) = 0.773

@1 = 39.38 deg

F31 = 0.00241 ( cos(180 + 39.38)i + sin(180 + 39.38)j) = - 0.00186 i - 0.00153 j

Consider force due to q2 on q3 is Fxi + Fyj

then Fnet = (Fx - 0.00186)i + (Fy - 0.00153)j

and j vector of Fnet is zero.

Fy = 0.00153 N

this force is toward +ve y axis that means q2 repel q3 so they are of same sign .

q2 is -ve charged.

angle of q2q3 with x axis. = 180 - @2

and cos@2 = (5.2^2 + 3.3^2 - 4.1^2) / (2 x 3.3 x 5.2) = 0.615

@2= 52.02 deg

hence Fy = (kq2 q3 / 3.3^2) sin(180 -52.02) = 0.00153

(9 x 10^9 x q2 x 3 x 10^-6 ) / (3.3^2) = 0.00153/sin(127.98)

q2 = 7.83 x 10^-7 C Or 0.783 uC

a) q2 = - 0.783 uC

b) Fnet = (Fx - 0.00186) = (0.00153 / sin127.98) cos(127.98) - 0.00186

Fnet = - 3.0545 x 10^-3 N

Magnitude = 3.05 x 10^-3 N Or 3.05 mN