Respuesta :
Answer:
The value of Q is - 8 C
Explanation:
Given;
the magnitude of first charge = 2 C
position of the first charge = 0
_ (2m) 0(+2C) +(2m)(Q)
------------------------------------------------------------------------------------
E₁ --------------------------------->
<------------------------------------------------------------------ E₂
E₁ and E₂ are equal in magnitude but opposite in direction
| E₂ | = | E₁ |
[tex]\frac{K*Q}{r^2} = \frac{K*2}{r^2} \\\\\frac{Q}{(4)^2} = \frac{2}{(2)^2}\\\\\frac{Q}{16} =\frac{2}{4}\\\\\frac{Q}{16} =\frac{1}{2}\\\\2Q = 16\\\\Q = 8 \ C[/tex]
Thus, since E₂ is opposite in direction to E₁, the Q = - 8 C
An electric charge has an electric field associated with it, and a moving electric charge produces a magnetic field. The value of Q will be -8 C.
What is an electric charge?
When the matter is held in an electric or magnetic field, it has an electric charge, which causes it to experience a force.
An electric charge has an electric field associated with it, and a moving electric charge produces a magnetic field.
The given data in the problem is;
The distance from + ve x = 2m
The distance from + ve y = 2m
A charge at the origin = +2 C
Both the electric field has the same megnitude but in the opposite direction. Hence the following relation can be applied.
|E₁|=E₂|
[tex]\rm \frac{Kq}{r^2} =\frac{K\times 2}{r^2} \\\\ \frac{Q}{r^2} = \frac{2}{2^2}\\\\ \frac{Q}{16} =\frac{1}{2}\\\\ 2Q=16 \\\\ \RM Q=8c[/tex]
Hence the value of Q will be -8 C.
To learn more about the electric charge refer to the link;
https://brainly.com/question/8163163
