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)

it doesn't. it was imagined up in the middle ages.

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.