The charges and coordinates of two charged particles held fixed in an x y plane

Question

The charges and coordinates of two charged particles held fixed in an xy plane are q1 = +2.8 C, x1 = 3.5 cm, y1 = 0.50 cm and q2 = 4.6 C, x2 = 2.0 cm, y2 = 1.5 cm.

a. Find the magnitude and direction of the electrostatic force on particle 2 due to particle 1.

(b) At what x and y coordinates should a third particle of charge q3 = +4.2 C be placed such that the net electrostatic force on particle 2 due to particles 1 and 3 is zero?

Thank you!

Explanation / Answer

given,

charge q1=2.8 uC and its position is (x1.y1)=(3.5, 0.5) cm =(0.035, 0.005) m

charge q2=-4.6 uC and its position (x2,y2)=(-2, 1.5) cm =(-0.02, 0.015) m

a)

force on q2 is,

F=k*q1*q2/r^2

here,

r^2=(x2-x1)^2+(y2-y1)^2

r^2=(-0.02-0.035)^2+(0.015-0.005)^2

r^2=31.25*10^-4 m^2

r=0.056 m

now,

F=k*q1*q2/r^2

F=(9*10^9)*(2.8*4.6*10^-12)/(31.25*10^-4)

magnitude of force F=37.1 N

and

tan(theta)=(y2-y1)/(x2-x1)

tan(theta)=(0.015-0.005)/(-0.02-0.035)

===>

theta=10.3 degrees

b)

let third charge is q3=4.2 uC and its positon is (x3,y3)

now,

force between q1 and q2 = force between q2 and q3

F12=F23

k*q1*q2/r^2=k*q2*q3/r23^2

=====>

r23=r*sqrt(q3/q1)

r23=0.056*sqrt(4.3/2.8)

r23=0.0694 m =6.94 cm

and

x3=x2+r23*cos(180-theta)

=(-2 +6.94*cos(180-10.3)) = -8.8 cm

and

y3=y2+r23*sin(180-theta)

=1.5+6.94*sin(180-10.3) =2.7 cm

therefore position of charge q3 is (x3, y3)=(-8.8, 2.7) cm

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