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

Answer:

Explanation:

a)

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

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)

$F_{BC}=\frac{1}{4\pi\epsilon _o}\dot\frac{q_Bq_C}{r^2}\\\\=(8.98\times 10^9)\times \frac{(-6.12\times 10^{-4})(1.10\times 10^{-4})}{4^2+3^2}\\\\=23.70N(-ve)$

d)

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

e)

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

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)

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

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

from +ve x axis counter clockwise.