Question

Calculate the magnitude and direction of the Coulomb force on each of the three charges shown in the figure below 

A positive charge q1=3.20μC on a frictionless horizontal surface is attached to a spring of force constant k as in the figure shown below. When a charge of q2=8.650 μC is placed 9.50 cm away from the positive charge, the spring stretches by 5.00 mm, reducing the distance between charges to d 9.00 cm. 

Answer 1

here,

q1 = 6 uC = 6 * 10^-6 C

q2 = 1.5 uC = 1.5 * 10^-6 C

q3 = - 2 uC = - 2 * 10^-6 C

r1 = 3 cm = 0.03 m

r2 = 2 cm = 0.02 m

the magnitude of force on q1 , F1 = K * q1 * q3 /(r1 + r2)^2 - K * q1 * q2 /r1^2

F1= 9 * 10^9 * 6 * 10^-6 * 10^-6 * ( 2/(0.03 + 0.02)^2 - 1.5/(0.03^2)

F1 = 46.8 N

the direction of force is in left direction

the magnitude of force on q2 , F2 = K * q2 * q3 /( r2)^2 + K * q1 * q2 /r1^2

F2 = 9 * 10^9 * 1.5 * 10^-6 * 10^-6 * ( 2/( 0.02)^2 + 6/(0.03^2)

F2 = 157.5 N

the direction of force is in right direction

the magnitude of force on q3 , F3 = K * q1 * q3 /(r1 + r2)^2 + K * q3 * q2 /r2^2

F3 = 9 * 10^9 * 2 * 10^-6 * 10^-6 * ( 6/(0.03 + 0.02)^2 + 1.5/(0.02^2)

F3 = 110.7 N

the direction of force is in left direction

Answer 2

SOLUTION : (one problem at a time is answered) :

FIRST PROBLEM :

Let charges be named as q1, q2 and q3 from left to right.

Distance between them be named as d12 , d23 and d13.

Movement to the left as positive force and to the right as negative.

For charge q1 :

Force on charge q1 by q3 (attracting q1 to the right), F13

= k *q1 * q3 / (d13)^2 

= 9 * 10^9 * 6.00 * 10^(-6) * (-2.00) * 10^(-6) / (3.00 + 2.00)^2  * 10^4  

( as k = 9*10^9 N m^2/C^2)

= - 43.2 N 

Force on charge q1 by q2 (repelling q1 to the left), F12

= k *q1 * q2 / (d12)^2 

= 9 * 10^9 * 6.00 * 10^(-6) * 1.50 * 10^(-6) / (3.00)^2  * 10^4

= 90  N 

So,

Net force on q1 

= - 43.2 + 90

= 46.8 N (ANSWER). 

Direction of this force = to the left . (ANSWER) 

For charge q2 :

Force on q2 by q1 =  - 90 N (repelling to the right) 

Force on q2 by q3 (attracting to the right)

= k q2 q3 / (r23)^2

=  9*10^9 * (1.5)*10^(-6) * (-2.00)*10^(-6) / (- 2.00)^2 *10^4

= - 67.5 N

So,

Net  force on q2 = - 90 - 67.5 = - 157.5 N 

Magnitude of force on q2 = 157.5 N (ANSWER).

Direction : to the right (ANSWER)

For q3 :

Force on q3 by q1 and q3

= 43.2 + 67.5

= 110.7 N (ANSWER) 

Direction to the left (as force has poise value)  (ANSWER)