Answer:
The charge 3q must be placed at a distance of 1.73d away from the origin.
Step-by-step explanation:
The electric field at the origin due to q at distance, d from the origin is E = q/4πε₀d².
The electric field at the origin due to 2q at distance, d from the origin is E' = -2q/4πε₀d². (It is negative since it its directed towards the negative x-axis)
Now, the net electric field at the origin due to both charges is thus E" = E + E' = q/4πε₀d² + (-2q/4πε₀d²) = q/4πε₀d² - 2q/4πε₀d² = -q/4πε₀d².
The electric field due to 3q at a distance x from the origin is E₁ = 3q/4πε₀x².
Now, the net electric field E' must cancel out E₁ so that the electric field at the origin is zero.
So, E' + E₁ = 0
E' = -E₁
-q/4πε₀d² = -3q/4πε₀x²
1/d² = 3/x²
taking inverse of both sides, we have
x²/3 = d²
multiplying both sides by 3, we have
x² = 3d²
taking square root of both sides, we have
x = (√3)d
x = ±1.73d
So, the charge 3q must be placed at a distance of 1.73d away from the origin.