A particle with charge q on the negative x axis and a second particle with charge 2q on the positive x axis are each a distance d from the origin. Where should a third particle with charge 3q be placed so that the magnitude of the electric field at the origin is zero? (Use any variable or symbol stated above as necessary.)

The charge 3q must be placed at a distance of 1.73d away from the origin.

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.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be