# Physics Electric Field Question?

A particle with charge 2q on the negative x axis and a second particle with charge 4q on the positive x axis are each a distance d from the origin. Where should a third particle with charge 6q be placed so that the magnitude of the electric field at the origin is zero? (Use any variable or symbol stated above as necessary.)

Relevance

Find the resultant field at the origin and then place 6q to exactly counter it in magnitude and direction

At the origin E = kq(2/d² - 4/d²) = -2kq/d² = k2q/d² at 180°

k6q/a² = k2q/d²

6d² = 2a² so a =√3*d so the 6q must be √3 *d left of the origin

or at x = -√3 *d <<<<<<

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• A particle with charge 2q on the negative x axis and a second particle with charge 4q on the positive x axis are each a distance d from the origin. Where should a third particle with charge 6q be placed so that the magnitude of the electric field at the origin is zero? (Use any variable or symbol stated above as necessary.)

assuming all charges are +

let position of charge 3 be D, on the –x axis.

field due to first charge E₁ = k2q/d² pointed right

field due to 2nd charge E₂ = k4q/d² pointed left

field due to 3rd charge E₃ = k6q/D² pointed right

sum equals 0

k2q/d² + k6q/D² = k4q/d²

1/d² + 3/D² = 2/d²

3/D² = 1/d²

D² = 3d²

D = d√(3)

Electric field

The strength or magnitude of the field at a given point

is defined as the force that would be exerted on a

positive test charge of 1 coulomb placed at that point;

the direction of the field is given by the direction of

that force.

E = F/Q = kQ/r²

Q = F/E F = QE

in Newtons/coulomb OR volts/meter

k = 1/4πε₀ = 8.99e9 Nm²/C²

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