Question

7. Calculate the force on the charge at the origin due to the other two charges (ollow the steps below). Different approach than Prob. 3, same answer at the end. q+ 2m h) Calculate the magnitude of the electric field of the 2nd charge at the origin E2 = N/C f the 3rd charge at the origin E3 Sketch arrows on the diagram above at the origin to indicate the vector a N/C rrows of E2 and j) Put the components of each vector in the table below to get the components of the resultant k) E field x-comp y-comp 1) Use the Pythagorean theorem. to get R=Je +R- m) Find the angle, θ = tan-11 n) Sketch an arrow representing the resultant on the figure above. o) Use the electric field calculated in l) and m) to get the resultant and magnitude on test N/C charge q' qi Nand θ = Compare answer with Prob. 3

Answer 1

7.

h)

q₂ = magnitude of second charge = 2 uC = 2 x 10^-6 C

r₂ = distance of second charge from the origin = 1 m

E₂ = magnitude of electric field by second charge at origin

E₂ = electric field by the second charge at the origin = k q₂/r₂² = (9 x 10⁹) (2 x 10^-6)/(1)² = 18000 N/C

i)

q₃ = magnitude of third charge = 3 uC = 3 x 10^-6 C

r₃ = distance of third charge from the origin = 2 m

E₃ = magnitude of electric field by third charge at origin

E₃ = electric field by the third charge at the origin = k q₂/r₂² = (9 x 10⁹) (3 x 10^-6)/(2)² = 6750 N/C

j)

the electric field are directed at the origin like this

k)

L)

Rx = 6750 N/C

Ry = - 18000 N/C

using Pythagorean theorem

R = sqrt(R_x² + R_y²) = sqrt((6750)² + (- 18000)²)

R = 19224 N/C

m)

= tan^-1(R_y/R_x) = tan^-1(- 18000/6750) = - 69.5 deg

negative sign indicate the angle below the positive x-axis

n)

o)

F₁ = q₁ R

F₁ = (1 x 10^-6) (19224)

F₁ = 0.019224 N

= 69.5 deg