Question

Three point charges are placed on the x-axis. A charge of +2.0 μC is placed at the origin, -2.0 μC to the right at x = 50 cm, and +4.0 μC at the 100 cm mark. What are the magnitude and direction of the electrostatic force which acts on the charge at the origin?

Answer 1

charge on q₁ = +2 x 10^-6 C                                           (at x = 0 m)

charge on q₂ = -2 x 10^-6 C                                            (at x = 0.5 m)

charge on q₃ = +4 x 10^-6 C                                           (at x = 1 m)

According to the coulomb's law,

Force between charge q₁ & charge q₂ :

F₁ = k q₁ q₂ / r₁² { eq.1 }

Force on charge q1 due to q₃ :

F₂ = k q₂ q₃ / r₂² { eq.2 }

The magnitude of the electrostatic force which acts on the charge at the origin which is given as ::

F_net = F₁ - F₂

F_net = k q₁ q₂ / r₁² - k q₁ q₃ / r₂²                                       { eq.3 }

where, k = 9 x 10⁹ Nm²/C²

r₁ = 0.5 m   &   r₂ = 1 m

inserting all these values in eq.3,

F_net = (9 x 10⁹ Nm²/C²) (+2 x 10^-6 C) (-2 x 10^-6 C) / (0.5 m)² - (9 x 10⁹ Nm²/C²) (+2 x 10^-6 C) (+4 x 10^-6 C) / (1 m)²

ignoring the negative sign.

F_net = (144 x 10^-3 N) - (72 x 10^-3 N)

F_net = 72 x 10^-3 N

Direction of the electrostatic force will be given as ::

Towards the right