Question

Particle A of charge 3.15 10-4 C is at the origin, particle B of charge -6.12 10-4 C is at (4.00 m, 0), and particle C of charge 1.10 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.(a) What is the x component of the electric force exerted by A on C?

N
(b) What is the y component of the force exerted by A on C?
N
(c) Find the magnitude of the force exerted by B on C.
N
(d) Calculate the x component of the force exerted by B on C.
N
(e) Calculate the y component of the force exerted by B on C.
N
(f) Sum the two x components from parts (a) and (d) to obtain the resultant x component of the electric force acting on C.
N
(g) Similarly, find the y component of the resultant force vector acting on C.
N
(h) Find the magnitude and direction of the resultant electric force acting on C.
magnitude N
direction � counterclockwise from the +x-axis

Answer 1

Step-by-step explanation:

a)

`F_(AC)=(1)/(4\pi\epsilon _o)\dot(q_Aq_C)/(r^2)\\\\=(8.98* 10^9)* ((3.15* 10^(-4))(1.10* 10^(-4)))/(3^2)\\\\=33.88N\vec{j}`

There will be no x-component as it is acting on y-axis completely

b)

y-component will be the total force
`F_(AC)=33.88N\vec j` as found earlier

c)

`F_(BC)=(1)/(4\pi\epsilon _o)\dot(q_Bq_C)/(r^2)\\\\=(8.98* 10^9)* ((-6.12* 10^(-4))(1.10* 10^(-4)))/(4^2+3^2)\\\\=23.70N(-ve)`

d)

x-component of
`F_(BC)=F_(BC)cos\theta\\\\=23.70((4)/(5))\\\\=18.96N\vec{i}`

e)

y-component of
`F_(BC)=F_(BC)sin\theta\\\\=23.70((3)/(5))\\\\=14.22N\vec{-j}`

f)

`F_x`=(x-component of
`F_(AC))+(x-component of [tex]F_(BC)`)
`=0 + 18.96\vec i=18.96\vec i`

g)

`F_y=(33.88)+(-14.22)\\\\=19.66\vec {j}`

h)

`F=F_x+F_y\\\\=18.96\vec i+19.66\vec j`

`|F|=√((18.96)^2+(19.66)^2)\\\\=27.31N\\\\tan\alpha=(F_y)/(F_x)\\\\=frac{19.66}{18.96}\\\\\alpha=46.04^o`

from +ve x axis counter clockwise.