Two fixed particles, each of charge 5.6 ✕ 10−6 c, are 24 cm apart. what force (in n) do they exert on a third particle of charge −2.6 ✕ 10−6 c that is 13 cm from each of them? (express your answer in vector form. assume that the +x-axis is to the right and the +y-axis is up along the page. further assume that the two particles sit along the x-axis and the third particle sits above the x-axis.)

Vector Fn₃ : Net force exerted by q₁ and q₂ on q₃

Fn₃=(0(i)+5.16j ) N  

Fn₃= 5.16 N (+y)

Explanation:

To solve this problem we apply Coulomb's law:

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:

o solve this problem we apply Coulomb's law:  

Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:  

F=K*q₁*q₂/d² Formula (1)  

F: Electric force in Newtons (N)

K : Coulomb constant in N*m²/C²

q₁,q₂:Charges in Coulombs (C)  

d: distance between the charges in meters(m)

Equivalence  

1nC= 10⁻⁹C

1cm=10⁻²m

Data

K=9x10⁹N*m²/C²

q₁ =q₂= 5.6nC = +5.6 *10⁻⁹C

d₁=d₂=13cm=13*10⁻²m= 0.13 m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on  q₃ are shown in the attached figure.

F₁₃:Force exerted by q₁ over q₃: attractive because the charges have opposite signs.

F₂₃:Force exerted by q₂over q₃: attractive because the charges have opposite signs.

y₃ : vertical position of q₃

We use formula (1) to calculate the electric forces:

F₁₃=F₂₃=The magnitude of F₁₃ is equal to the magnitude of F₂₃ because they have the same charge and the same distance (d)

F₁₃=F₂₃=9*10⁹*5.6*10⁻⁶*2.6*10⁻⁶/0.13²= 7.75 N

Magnitudes, direction and sense of the x-y components of F₁₃ and F₂₃

F₁₃x= F₂₃x   , F₁₃x( in direction -x)  ,   F₂₃x: ( in direction +x )

F₁₃y=F₂₃y    , F₁₃y( in direction +y)  ,   F₂₃y:( in direction +y)

x-y components of the net force in q₃ (Fn₃)

Fn₃x= - F₁₃x+ F₁₃x =0

Fn₃y = F₁₃y +F₂₃y

F₁₃y=F₂₃y= F₁₃ sinβ=7.75*(5÷15) =2.58 N

Fn₃y = 2.58 N+2.58 N= 5.16 N

Vector Fn₃

Fn₃=(0(i)+5.16j ) N  

Fn₃= 5.16 N (+y)

a displacement distance is size of displ

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer