**Answer:**

N: Resulting force on Q in the direction of the x axis

N: Resulting force on Q in the direction of the y axis

**Explanation:**

To solve this problem we need 3 basic concepts:

1.**Coulomb's law:
**

Two point charges (q1, q2) separated by a distance (r) exert a mutual force (F) whose magnitude is determined by the following formula:

Formula (1)

: Coulomb constant

q 1,q2=charge in Coulombs (C)

The direction of F is determined by the unit vector ( ) in the direction of F

: position vector in the direction of F

r= (x2-x1)i+( y2-y1)j

The points (x1, y1) and (x2, y2) represent the position coordinates of the charges
.

The sense of force depends on the sign of q1 and q2:

When the charges have opposite signs the force is attractive.

When the charges have an equal sign, the force is repulsion

2. **Newton's third law: law of action and reaction
**

The magnitude of the force exerted by q1 on q2 is equal to the force exerted by q2 on q1, and these forces have the opposite direction.

<h3>F1-2=-F2-1
</h3>

**Problem development**

We have the forces exerted by the charges q1 + and q2 + on the load Q-.

The forces F1 and F2 on the load Q are attractive forces, Because of this the force comes out of Q and is directed towards q1 and q2

**Calculation of vector r1 **: points Q(0.4,0.15) , q1(0,0.3)

r1= (0- 0.4)i+(0.3-1.5)j

r1=-0.4i+0.15j

magnitude of r1

**Calculation of vector r2** : points(0.4,0.15) (0,-0.3)

r2= (0- 0.4)i+(-0.3-1.5)j

r2=(-0.4i-0.45j) m

magnitude of r2

We apply formula 1 to calculate the magnitudes of F1 and F2 with the known data :

q1 = q2 = 2μC , r1= 0.1825m ,r2=0.6m,

1 μC=

Calculation of the total electric force (F) exerted by the charges q1 and q2 on the charge Q

: unit vector in the direction of r

ru=(rx+ry)/r

F=(-473.27i+177.47j-13.31i-14.98j) ))*10e-2 N

Answer:

N: Resulting force on Q in the direction of the x axis

N: Resulting force on Q in the direction of the y axis

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