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The proton (positive charge) shall be closer to the charge with the lower magnitude which is at the orgin.
The proton also shall be out of the interval between the two charges so that the pull of one charge cancels with the push of the other.
The region at which those conditions happen is to the left of the origin.
In that case the forces over the proton shall be:
k* (2.4 nC) * p / (x^2) - k*(4.8 nC) * p /( 1.3 + x)^2 = 0
where p is the charge of the proton.
You can simplify k and p:
2.4 / x^2 - 4.8 / (1.3 + x)^2 = 0
You can also simplify by 2.4
1/ x^2 - 2 / (1.3 + x)^2 = 0
(1.3+x)^2 - 2x^2 = 0
1.69 + 2.6x + x^2 - 2x^2 = 0
1.69 + 2.6x - x^2 = 0
x^2 -2.6x - 1.69 =0
Solve using the quadratic formula: x = 3.14 (use only the positive value)
That is the proton shall be place 3.14 units to the left of the origin (positive charge)
the answer is in the picture ok
good luck don't worry it's a correct
#carry on learning
Answer:
At centroid
Explanation:
In the given equilateral triangle ABC with side of 1 m. The three equal charges e,e,e are placed at the A,B and C.
And the fourth charge 2e is put at point O which is called centroid.
Now we can calculate the distance AD by applying pythagorean theorem as,
Put the values and get.
Now calculate AO as,
And the sides BO=CO=AO.
Now Force can be calculated as
And similarly,
Now we can calculate resultant of in upward direction. as,
Therefore the resultant force on centroid O.
Therefore the fourth charge 2e should be placed on centroid so that it experience zero force.