Answer:
x = 0, y = 1.5 m
Explanation:
The magnitude of the electric field strength is given by the formula
E=[tex]\frac{Kq }{r}[/tex]
Where
E is the magnitude of the electric field at the a point in space
k is the universal Coulomb constant
q is the charge of the particle
r is the distance that the point in space, at which we want to know the E, is from the point charge that is causing E
Two positive charges in the question can be referred to as q₁ and q₂.
q₁ =6μC
q₂ =4μC
Each of the positive charges exert electric field of magnitude of E₁ and E₂ respectively. For positive charge, the electric field direction is away from the positive charge, unlike in negative charge where the electric field has a direction towards the negative charge. The resultant magnitude of the electric field exerted by both positive charges is E₀.
E₀=E₁ +E₂
Since the direction of the electric field is outward from each of the positive charge q1 and q2, their electric field strength will be opposite, hence
E₀ = E₁ - E₂
From the question, we were ask to find point where E₀ is zero, so E₁=E₂. Using the origin as point of reference, let the point where the electric field strength is zero x from position of q₁ and 1-x from q₂. So that
E₁ =[tex]\frac{kq_{1} }{x^{2} }[/tex] and E₂ =[tex]\frac{kq_{2} }{(1-x)^{2} }[/tex]
Now E₁=E₂
[tex]\frac{kq_{1} }{x^{2} }[/tex] =[tex]\frac{kq_{2} }{(1-x)^{2} }[/tex]
Divide both sides of the equation by k
[tex]\frac{q_{1} }{x^{2} }[/tex] =[tex]\frac{q_{2} }{(1-x)^{2} }[/tex]
q₁(1-x)²=q₂x²
q₁*(1 - 2x - x²) = q₂x
Remember that q₁ = 6μC and q₂ = 4μC
6(1 - 2x - x²) = 4x²
3(1 - 2x - x²) = 2x²
3 - 6x - 3x² = 2x²
3 - 6x - 5x² = 0
This is a Quadratic Equation of the form ax² + bx + c
x = [tex]\frac{-b +/- \sqrt{b^{2}-4ac } }{2a}[/tex]
From our quadratic equation 3 - 6x - 5x² = 0
a = -5 , b = -6 and c = 3
x = [tex]\frac{-(-6) +/- \sqrt{(-6)^{2} -(4*-5*3)} }{2*-5}[/tex]
x = [tex]\frac{6 +/- \sqrt{36 - (-60)} }{-10}[/tex]
x= [tex]\frac{6 +/- 9.792}{-10}[/tex]
x = [tex]\frac{15.797}{-10}[/tex] or [tex]\frac{-3.797}{-10}[/tex]
x = -1.5 or 0.379
The negative sign in -1.5 simply indicate the direction of the electric field. Note that since both charges are the same, there would be repulsion which is likely going to increase the distance between the two positive charges. Therefore the correct point where the electric field strength is zero is 1.5 m along the y-axis.
