Question

A charge of -3.35 nC is placed at the origin of an xy-coordinate system, and a charge of 1.75 nC is placed on the y axis at y = 3.90 cm . A. If a third charge, of 5.00 nC , is now placed at the point x = 2.70 cm , y = 3.90 cm find the x and y components of the total force exerted on this charge by the other two charges.

B. Find the magnitude of this force.
C. Find the direction of this force. ( ° below the +x axis )

Answer 1

Step-by-step explanation:

Force due to charges 1.75 and 5 nC is given below

F =K Q₁Q₂ / d²

F₁ =
`(9*10^9*1.75*10^(-9)*5* 10^(-9))/((2.7*10^(-2))^2)`

F₁ = 10.8 X 10⁻⁵ N . It will at in x direction.

Force due to other charge placed at origin

F₂ =
`(9*10^9*1.75*10^(-9)*5* 10^(-9))/(22.5*10^(-4))`

F₂ = 3.5 x 10⁻⁵ N.

Its x component

= F₂ Cos θ

= 3.5 x 10⁻⁵ x 3.9/ 4.74

= 2.88 x 10⁻⁵ N

Its y component

F₂ sin θ

= 3.5 x 10⁻⁵ x 2.7/4.743

= 1.99 x 10⁻⁵ N

Total x component

= 10.8 X 10⁻⁵ +2.88 x 10⁻⁵

= 13.68 x 10⁻⁵ N.

Magnitude of total force F

F² = (13.68 x 10⁻⁵)² + (1.99 x 10⁻⁵ )²

F = 13.82 X 10⁻⁵ N

Direction θ with x axis .

Tanθ = 1.99/ 13.68

θ = 8 °