Question

In the figure what are the (a) magnitude and (b) direction (from x-axis in the counterclockwise direction) of the net electrostatic force on particle 4 due to the other three particles? All four particles are fixed in the xy plane, and q_1 = -3.20 times 10^-19 C, q_2 = +3.20 times 10^-19 C, q_3 = +6.40 times 10^-19 C, q_4 = +3.20 times 10^-19 C, theta_1 = 42 degree, d_1 = 6.20 cm, and d_2 = d_3 = 2.90 cm.

Answer 1

Force is given by F=K(q1q2)/r²

Force on 4 due to 1 F₁=(9 $\times$ 10⁹) $\times$ (3.20 $\times$ 10^-19)²/ (0.0620)² (Here magnitude of q1 and q2 is same)

F₁=2.4 $\times$ 10^-25 N this attractive force acting towards 1.

The x component of the force F_1x=F₁cos42⁰

= (2.4 $\times$ 10^-25)cos42⁰

F_1x= 1.8 $\times$ 10^-25 N along -X axis

The Y componet of the force F_1y=F₁ sin42⁰

   = (2.4 $\times$ 10^-25)sin42⁰

= 1.6 $\times$ 10^-25   N along -Y axis .

Now force on 1 due to charge 2

F₂=(9 $\times$ 10⁹) $\times$ (3.20 $\times$ 10^-19)²/ (0.0290)²

F₂=11.0 $\times$ 10^-25 N repulsive force acting along -Y axis.

Force on 4 due to 3

F₃=(9 $\times$ 10⁹) $\times$ (3.20 $\times$ 10^-19 $\times$ 6.4 $\times$ 10^-19)/ (0.0290)²

F₃=22.0 $\times$ 10^-25 N Acting along -X axis .

Now all forces along X axis F_x=1.8 $\times$ 10^-25 +22.0 $\times$ 10^-25 Both component is in same direction -X axis.

(there is no component of force in X direction due to charge 2 )

F_x=23.8 $\times$ 10^-25 N

Now all forces along Y axis F_y= 1.6 $\times$ 10^-25 +11.0 $\times$ 10^-25  

=12.6 $\times$ 10^-25 N along -Y axis

Now net force F=sqrt (F_x²+F_y²)

= sqrt ((2.38 $\times$ 10^-24)²+(1.26 $\times$ 10^-24)²)

F =2.70 $\times$ 10^-24 N ANS

Direction is $\theta$ =tan^-1(F_y/F_x)

=tan^-1((12.6 $\times$ 10^-25)/(23.8 $\times$ 10^-25))

   $\theta$ =27.9 ^{0   ANS}