Question

A particle (charge = -15.0 µC) is located on the x- axis at the point x = -25.0 cm, and a second particle (charge = +45.0 µC) is placed on the x- axis at x = +30.0 cm. What is the magnitude of the total electrostatic force on a third particle (charge = -3.50 µC) placed at the origin (x = 0)?

Answer 1

Given,

charge of first particle = –15.0μC = –15.0 x 10^–6 C

charge of the second particle = +45.0μC = +45.0x 10^–6 C

charge of the third particle = –3.5μC = –3.5 x 10^{–​​​​​​​6} C

distance of the first charge from the third charge =  25 cm = 25x10^–2m

distance of the second charge from the third charge = 30 cm = 30x10^–2m

Formula for electrostatic force between two charges = (1/4πε_o)x(q₁q₂/r²) = 9x10⁹x(q₁q₂/r²)

( 1/4πε_o = 9x10⁹Nm²/C²)

The force of repulsion on third charge due to first charge, F₁ = 9x10⁹ x (15x 10^{–​​​​​​​6} x 3.5 x 10^{–​​​​​​​6}) /(25x10^–2)²

= 9x10⁹ x (52.5 x 10^{–​​​​​​​12}) /(625x10^–4)

= 472.5x10 / 625

= 4725/625

F₁ = 7.56 N

The force of attraction on third charge due to second charge, F₂ = 9x10⁹ x (45x 10^{–​​​​​​​6} x 3.5 x 10^{–​​​​​​​6}) /(30x10^–2)²

= 9x10⁹ x (157.5^​ x 10^{–​​​​​​​12}) /(900x10^–4)

= 14170/900

F₂ = 15.74 N

Both F₁ and F₂ are acting in the same direction, towards positive-x direction.

So, the net force on the third charge due to first and second charges, F = F₁+ F₂ = 7.56+15.74 = 23.3N

So, the total electrostatic force on the third charged particle due first and second charged particles, is 23.3N along the positive x direction.