Three point charges are placed on the x-axis. A charge of +2.0 ¦ÌC is placed at the origin, −2.0 ¦ÌC to the right at x=50 cm, and +4.0 ¦ÌC at the 100cm mark. What is the magnitude of the electrostatic force that acts on the charge at the origin? (k=1/4¦Ðϵ0=8.99¡Á109 N ¡¤ m2/C2)

To find the magnitude of the electrostatic force that acts on the charge at the origin, we can use Coulomb's Law. Coulomb's Law states that the force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's Law is:

F = k * |q1 * q2| / r^2

Where:
F is the electrostatic force
k is Coulomb's constant (k = 1 / (4 * π * ϵ0) = 8.99 * 10^9 N * m^2 / C^2)
q1 and q2 are the charges
r is the distance between the charges

In this case, the charge at the origin is +2.0 µC, and the charges at x = 50 cm and x = 100 cm are -2.0 µC and +4.0 µC respectively.

First, let's calculate the distance between the charge at the origin and the charge at x = 50 cm:
r1 = 50 cm - 0 cm = 50 cm

Next, let's calculate the distance between the charge at the origin and the charge at x = 100 cm:
r2 = 100 cm - 0 cm = 100 cm

Now, we can calculate the magnitude of the electrostatic force on the charge at the origin by considering the forces between each pair of charges.

Force between the charge at the origin and the charge at x = 50 cm:
F1 = k * |q1 * q2| / r1^2
= (8.99 * 10^9 N*m^2/C^2) * |(2.0 * 10^-6 C) * (-2.0 * 10^-6 C)| / (0.50 m)^2

Force between the charge at the origin and the charge at x = 100 cm:
F2 = k * |q1 * q3| / r2^2
= (8.99 * 10^9 N*m^2/C^2) * |(2.0 * 10^-6 C) * (4.0 * 10^-6 C)| / (1.00 m)^2

Finally, to find the total electrostatic force on the charge at the origin, we sum up the forces:
F_total = F1 + F2