Question

A charge 5.05 nC is placed at the origin of anxy-coordinate system, and a charge -1.97 nC is placed on the positive x-axis at x = 3.99 cm . A third particle, of charge 6.03 nC is now placed at the point x = 3.99 cm , y = 3.00 cm .

(1)Find the x-component of the total force exerted on the third charge by the other two.

(2)Find the y-component of the total force exerted on the third charge by the other two.

(3)Find the magnitude of the total force acting on the third charge.

(4)Find the direction of the total force acting on the third charge.

Answer 1

Given that :

charge on q₁ = 5.5 x 10^-9 C                                                     at x = 0

charge on q₂ = -1.97 x 10^-9 C                                                 at x = 3.99 cm

charge on q₃ = 6.03 x 10^-9 C                                                  at x = 3.99 cm & y = 3 cm

Distance of q₃ from q₁, r₃₁ = (3.99 cm)² + (3 cm)² = 4.99 x 10^-2 m

Distance of q₃ from q₂, r₃₂ = 3 x 10^-2 m

The force exerted on q₃ due to q₁ which is given as :

F₃₁ = k q₃ q₁ / (r₃₁)²                                                                                                 { eq.1 }

F₃₁ = (9 x 10⁹ Nm²/C²) (6.03 x 10^-9 C) (5.5 x 10^-9 C) / (4.99 x 10^-2 m)²

F₃₁ = 11 x 10^-5 N                              (along a positive axis)

using an identity, tan = (3 cm) (3.99 cm)

= tan^-1 (0.7518)

= 36.94 degree

component along x axis, F_X = F Cos                                                                                 { eq.2 }

inserting the values in eq.2,

F_X = (11 x 10^-5 N) Cos (36.94⁰)

F_X = 7.97 x 10^-5 N (i)

component along y axis, F_Y = F Sin                                                                                        { eq.3 }

inserting the values in eq.3,

F_y = (11 x 10^-5 N) Sin (36.94⁰)

F_y = 6.6 x 10^-5 N (j)

Now, we have    F₃₁ = F_X + F_Y { eq.4 }

inserting the values in eq.4,

F₃₁ = [7.97 x 10^-5 N (i) + 6.6 x 10^-5 N (j)]

The force exerted on q₃ due to q₂ which is given as :

F₃₂ = k q₃ q₂ / (r₃₂)²                                         { eq.5 }

F₃₂ = (9 x 10⁹ Nm²/C²) (6.03 x 10^-9 C) (-1.97 x 10^-9 C) / (3 x 10^-2 m)²

F₃₂ = 11.8 x 10^-5 N (-j)                       (along a negative axis)

Total force, F_total = F₃₁ + F₃₂                                                                                                             { eq.6 }

then        F_total = [7.97 x 10^-5 N (i) + 6.6 x 10^-5 N (j)] + [11.8 x 10^-5 N (-j)]

F_total = 7.97 x 10^-5 N (i) - 5.2 x 10^-5 N (-j)   

(a) the x-component of the total force exerted on the third charge by the other two which is given as ::

F_x = 7.97 x 10^-5 N                                           (in +ve x-direction)

(b) the y-component of the total force exerted on the third charge by the other two which is given as :

F_y = 5.2 x 10^-5 N                                              (in -ve y-direction)

(c) magnitude of the total force acting on the third charge will be given as :

F_net = (7.97 x 10^-5 N)² + (5.2 x 10^-5 N)²

F_net = 90.56 x 10^-10

F_net = 9.51 x 10^-5 N

(4) direction of the total force acting on the third charge will be given as :

= tan^-1 [(5.2 x 10^-5 N) / (7.97 x 10^-5 N)]

= tan^-1 (0.6524)

= 33.12 degree