A 2μC charge is positioned at x = -0.8m and another 2μC charge is positioned at x = 0.8m. A positive test charge of q = 1.28*10^-18C is positioned at the origin x = 0m. Find (a) the electric force exerted on q, (b) the electric field at the origin, and (c) the electric potential at the origin.

My work and calculations:

(a) ∑F = -F1 + F2
= -(kq1q/r^2) + (kq2q/r^2)
= kq((-q1/r^2)+(q2/r^2))

I noticed that the terms in the brackets cancel out to 0. But that would result in ∑F = 0N. I'm not sure if I'm doing this correctly... Could someone please help me?

it is zero. Think..you have two positive charges equidistant on opposite sides...

To find the electric force exerted on the positive test charge at the origin, you need to calculate the forces exerted by each of the charges, and then add them together using vector addition.

Given:
Charge 1 q₁ = 2μC = 2 * 10^(-6) C
Position of charge 1 x₁ = -0.8 m
Charge 2 q₂ = 2μC = 2 * 10^(-6) C
Position of charge 2 x₂ = 0.8 m
Test charge q = 1.28 * 10^(-18) C
To calculate the electric force exerted by each charge on the test charge, you can use Coulomb's law, which states that:

F = k * |q₁ * q₂| / r²

Where:
F is the magnitude of the force
k is the electrostatic constant (k ≈ 9 x 10^9 Nm²/C²)
q₁ and q₂ are the charges
r is the distance between the charges

(a) Electric force exerted on the test charge:
To calculate the net electric force, you need to consider the forces exerted by each of the charges at the origin. These forces will have the same magnitude, but opposite directions. So, you can write:

∑F = F₁ + F₂

where F₁ is the force exerted by charge 1 and F₂ is the force exerted by charge 2.

Using Coulomb's law, the equation for each force will be:

F₁ = k * |q₁ * q| / r₁²
F₂ = k * |q₂ * q| / r₂²

where r₁ and r₂ are the distances from the charges to the origin, which are simply the absolute values of their respective positions.

Substituting the given values:
k = 9 x 10^9 Nm²/C²
q₁ = 2 * 10^(-6) C
q₂ = 2 * 10^(-6) C
q = 1.28 * 10^(-18) C
r₁ = |-0.8 m| = 0.8 m
r₂ = |0.8 m| = 0.8 m

Now we can calculate the forces:

F₁ = (9 x 10^9 Nm²/C²) * (2 * 10^(-6) C) * (1.28 * 10^(-18) C) / (0.8 m)²
F₂ = (9 x 10^9 Nm²/C²) * (2 * 10^(-6) C) * (1.28 * 10^(-18) C) / (0.8 m)²

Please note that you should plug the values into a calculator to get the numerical answer.

After calculating F₁ and F₂, you can sum them up:

∑F = F₁ + F₂

This will give you the net electric force exerted on the test charge at the origin.

(b) Electric field at the origin:
The electric field, E, at a given point is defined as the force per unit positive charge acting on a positive test charge placed at that point. So, the electric field due to charge 1 at the origin can be calculated as follows:

E₁ = F₁ / q
E₂ = F₂ / q

where E₁ is the electric field due to charge 1 and E₂ is the electric field due to charge 2.

Substituting the calculated forces:

E₁ = F₁ / q
E₂ = F₂ / q

This will give you the electric field at the origin.

(c) Electric potential at the origin:
The electric potential, V, at a given point is defined as the electric potential energy per unit charge. To calculate the electric potential at the origin, you need to consider the contribution from each charge, and then sum them up.

The electric potential due to charge 1 at the origin can be calculated as follows:

V₁ = k * q₁ / r₁
V₂ = k * q₂ / r₂

where V₁ is the electric potential due to charge 1 and V₂ is the electric potential due to charge 2.

Substituting the given values:

V₁ = (9 x 10^9 Nm²/C²) * (2 * 10^(-6) C) / (0.8 m)
V₂ = (9 x 10^9 Nm²/C²) * (2 * 10^(-6) C) / (0.8 m)

After calculating V₁ and V₂, you can sum them up:

V = V₁ + V₂

This will give you the electric potential at the origin.