A) Four charges 4 × 109 C at (0 m, 0 m), 4 × 109 C at (1 m, 2 m), 7 × 109 C at (

Question

A) Four charges 4 × 109 C at (0 m, 0 m),
4 × 109 C at (1 m, 2 m), 7 × 109 C
at (4 m, 4 m), and 8 × 109 C at
(1 m, 3 m), are arranged in the (x, y) plane
as shown.

Find the magnitude of the resulting force on
the 4 nC charge at the origin. The Coulomb
constant is 8.98755 × 109 N · m2/C2.
Answer in units of N.

B) What angle does the resultant force on the
4 nC charge at the origin make with the
positive x axis? Quadrant I lies between 0
and 90, quadrant II between between 90and 180, etc. Answer in units of degrees.

Explanation / Answer

Let,

q = -4 nc at orogin
q1 = -8 nc at(1,3)

q2 = -4 nc at (1,-2)

q3 = 7 nc at (-4,-4)

F1 = k*q1*q/r1^2 = 9*10^9*8*10^-9*4*10^-9/(1^2+3^2) = 2.88*10^-8 N

angle made by F1 with -x axis, theta1 = tan^-1(3/1) = 71.57 degrees

F1x = -F1*cos(theta1) = -2.88*10^-8*cos(71.57) = -0.91*10^-8 N

F1y = -F2*sin(theta2) = -2.88*10^-8*sin(71.57) = -2.73*10^-8 N

F2 = k*q2*q/r2^2 = 9*10^9*4*10^-9*4*10^-9/(1^2+2^2) = 2.88*10^-8 N

angle made by F2 with -x axis, theta2 = tan^-1(2/1) = 63.43 degrees

F2x = -F2*cos(theta2) = -2.88*10^-8*cos(63.43) = -1.29*10^-8 N

F2y = F2*sin(theta2) = 2.88*10^-8*sin(63.43) = 2.57*10^-8 N

F3 = k*q2*q/r2^2 = 9*10^9*7*10^-9*4*10^-9/(4^2+4^2) = 0.78*10^-8 N

angle made by F3 with -x axis, theta3 = tan^-1(4/4) = 45 degrees

F3x = -F3*cos(theta3) = -0.78*10^-8*cos(45) = -0.55*10^-8 N

F3y = -F3*sin(theta3) = -0.78*10^-8*sin(45) = -0.55*10^-8 N

Fnetx = F1x + F2x + F3x = -2.75*10^-8 N
Fnety = F1y + F2y + F3y = -0.71*10^-8 N

Fnet = sqrt(Fnetx^2 + Fnety^2) = 2.84*10^-8 N <<<<<<---------Answer

direction ,

theta = tan^-1(Fnety/Fnetx)

= tan^-1(0.71/2.75)

= 14.47 below -x axis

or 194.47 degrees with +x axis in counter clsokwise direction <<<<<<---------Answer

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