Question

Question 1 18 pts 1) A charge, q1, of -9 micro-coulombs is located at the origin, a charge, q2, of -10 micro-coulombs is located at x = 0 cm, y = +9.9 cm, a charge, q3, of +8 micro- coulombs is located at x 0 cm, y #-3.2 cm, and a charge, q4, of +4 micro- coulombs is located at x = 9.7 cm, y = 0 cm, what is the magnitude of the total electric force on q2 (charge on the ty axis) in Newtons?

Answer 1

Force diagram for the given problem is shown below ;

Where, $F_{4}$ is force on the charge $q_{2}$ due to the charge $q_{4}$

  $F_{3}$  is force on the charge $q_{2}$ due to the charge $q_{3}$ ​

$F_{1}$ is force on the charge $q_{2}$ due to the charge $q_{1}$

Given, $q_{1}$ = - 9 x 10^-6 C located at origin

$q_{2}$ = -10 x 10^-6 C located at A (0 , 9.9 cm) or A (0 , 9.9 x 10^-2 m)

$q_{3}$ = 8 x 10^-6 C located at C (0 , -3.2 cm) or C (0 , -3.2 x 10^-2 m)

  $q_{4}$ = 4 x 10^-6 C located at B (9.7 cm , 0) or B (9.7 x 10^-2 m , 0)

The electrostatic force between two charge particles q and Q and separated by a distance r is given by

where,   

Using this formula we can calculate the magnitude of force acting on $q_{2}$ due to all other particles. (here i am only calculating the magnitude of forces. So, i will not consider the polarity of charges. The direction of force is already shown in the diagram).

where, OA = 9.9 x 10^-2 m

So,

In vector notation :

and

9342 1 (8 × 10-0) (10 x 10-0) F3 (OA +OC)2

Now, OA + OC = (9.9 + 3.2) x 10^-2 m = 13.1 x 10^-2 m

So,   

In vector notation

  

And finally

In the right angle triangle AOB , AB is the hypotenuse of the triangle.

So,,   

V 192.1 × 10-4 13.86× 10-2 m 7n

So,   

And, from the figure

  

or,  

So, in vector notation $F_{4}$ can be written as

or,   

or, (in N)

So, total electric force on $q_{2}$ can be given as

or,

or,

So, magnitude of total electric force on $q_{2}$ is

For any doubt please comment and please give an up vote. Thank you.