Question

A particle (charge = +40 µC) is located on the x axis at the point x = –20 cm, and a second particle (charge = –50 µC) is placed on the x axis at x = +30 cm. What is the magnitude of the total electrostatic force on a third particle (charge = –4.0 µC) placed at the origin (x = 0)?

Answer 1

Both of the first two charges push/pull the third (negative) to the left, and so there effects are additive:

F = Σ kQq / d² = 8.99e9N·m²/C² * 4.0e-6C * (40e-6C/(0.20m)² + 50e-6C/(0.30m)²)

F = 56 N ◄ magnitude

If that's not quite the correct interpretation, let me know.

Hope this helps!

Answer 2

Answer:

A particle (charge = +40 µC) is located on the x axis at the point x = –20 cm, and a second particle (charge = –50 µC) is placed on the x axis at x = +30 cm, the magnitude of the total electrostatic force on a third particle

(charge = –4.0 µC) placed at the origin (x = 0) is 16N

Explanation:

• The electrostatic force between two point charges ($q_{1} ,q_{2}$) separated by distance (r) is given by the formula

      $F=K\frac{q_{1} q_{2} }{r^{2} }$

      where the constant $K=9*10^{9}$

From the question, the charges are arranged as below

the force on the charge $Q=-4*10^{-6}C$ due to the first charge is

$F_{1} =\frac{9*10^{9}*40*10^{-6}*-4*10^{-6}}{0.2^{2} }$

⇒ $F_{1} =-36N$

force due to the second charge,

$F_{2} =\frac{9*10^{9}*-50*10^{-6}*-4*10^{-6}}{0.3^{2} }$

⇒$F_{2} =20N$

net force $F=-36+20=-16N$

the negative means that the force is repulsive