Question

There are only two charged particles in a particular region. Particle 1 carries a charge of 3q and is located on the negative x-axis a distance d from the origin. Particle 2 carries a charge of -2q and is located on the positive x-axis a distance d from the origin.
Required:
Where is it possible to have the net field caused by these two charges equal to zero?

Answer 1

Answer:

The net field will be the sum of the fields created by each charge.

where the charge Q in a position r' is given by:

E(r) = k*Q/(r - r')^2

Where k is a constant, and r is the point where we are calculating the electric field.

Then for the charge 3q, in the position r₁ = (-d, 0, 0) the electric field will be:

E₁(r) = k*3q/(r - r₁)^2

While for the other charge of -2q in the position r₂ = (d, 0, 0)

The electric field is:

E₂(r) = -k*2*q/(r - r₂)^2

Then the net field at the point r is:

E(r) = E₁(r) + E₂(r) = k*3q/(r - r₁)^2 + -k*2*q/(r - r₂)^2

E(r) = k*q*( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Then if the we want to find the points r = (x, y, z) such that:

E(r) = 0 = k*q*( 3/(r - r₁)^2 - -k*2*q/(r - r₂)^2)

Then we must have:

0 = ( 3/(r - r₁)^2 - 2/(r - r₂)^2)

Also remember that the distance between two points:

(x, y, z) and (x', y', z') is given by:

D = √( (x - x')^2 + (y - y)^2 + (z -z')^2)

Then we can rewrite:

r - r₁ = √( (x - (-d))^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x + d))^2 + y^2 + z^2)

and

r - r₂ =  √( (x - d)^2 + (y - 0 )^2 + (z -0)^2)

       = √( (x - d))^2 + y^2 + z^2)

Replacing that in our equation we get:

0 = ( 3/(√( (x + d))^2 + y^2 + z^2))^2 - -k*2*q/(√( (x - d))^2 + y^2 + z^2))^2)

0 = (3/((x + d))^2 + y^2 + z^2) - 2/ (x - d))^2 + y^2 + z^2)

We want to find the values of x, y, z such that the above equation is true.

2/ (x - d))^2 + y^2 + z^2) = (3/((x + d))^2 + y^2 + z^2)

2*[((x + d))^2 + y^2 + z^2] = 3*[(x - d))^2 + y^2 + z^2]

2*(x + d)^2  + 2*y^2 + 2*z^2 = 3*(x - d)^2 + 3*y^2 + 3*z^2

2*(x + d)^2 - 3*(x - d)^2 =  3*y^2 + 3*z^2 -  2*y^2 - 2*z^2

2*(x + d)^2 - 3*(x - d)^2  = y^2 + z^2

2*x^2 + 2*2*x*d + 2*d^2 -  3*x^2 + 3*2*x*d - 3*d^2 = y^2 + z^2

-x^2 + 10*x*d - d^2 = y^2 + z^2

we can rewrite this as:

- ( x^2 - 10*x*d + d^2) =  y^2 + z^2

now we can add and subtract 24*d^2 inside the parenthesis to get

- ( x^2 - 10*x*d + d^2 + 24*d^2 - 24*d^2) =  y^2 + z^2

-( x^2 - 2*x*(5d) + 25d^2 - 24d^2) = y^2 + z^2

-(x^2 - 2*x*(5d) + (5*d)^2) + 24d^2 = y^2 + z^2

The thing inside the parenthesis is a perfect square:

-(x - 5d)^2 + 24d^2 = y^2 + z^2

we can rewrite this as:

24d^2 = y^2 + z^2 + (x - 5d)^2

This equation gives us the points (x, y, z) such that the electric field is zero.

Where we need to replace two of these values to find the other, for example, if y = z = 0

24d^2 = (x - 5d)^2

√(24d^2)  = x - 5d

√24*d = x - 5d

√24*d + 5d = x

so in the point (√24*d + 5d, 0, 0) the net field is zero.