Question

A charge of -3.00 nC is placed at the origin of an x-y coordi- nate system, and a charge of 2.00 nC is placed on the y axis at y 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point x 3.00 cm, y 4.00 cm, find the x and y components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Answer 1

(a) Fnx = 6.76*10⁻⁵ N , Fny = - 4.3*10⁻⁵ N

(b) Fn= 8*10⁻⁵ N :magnitude of the total force exerted on q3

Direction of the total force exerted on q3 (α )

α = 32.46°, With respect to the positive axis of the x, in the fourth quadrant x-y

Step-by-step explanation:

Conceptual analysis

Because the particle q₃ is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

Equivalences

1nC= 10⁻⁹C

1cm = 10⁻²m

Known data

k= 9*10⁹N*m²/C²

q₁= -3 nC =-3*10⁻⁹C

q₂= +2.00 nC = 2*10⁻⁹C

q₃= +5.00 nC= =+5*10⁻⁹C

`d_(13) =\sqrt{3^(2)+4^(2)  } = 5cm = 0.05m`

d₂₃ = 3cm = 0.03 m

Graphic attached

The directions of the individual forces exerted by q₁ and q₂ on q₃ are shown in the attached figure.

The force (F₂₃) of q₂ on q₃ is repulsive because the charges have equal signs and the forces.

The force (F₁₃) of q₁ on q₃ is attractive because the charges have opposite signs.

Magnitudes of F₁₃ and F₂₃

F₁₃ = (k*q₁*q₃)/(d₁₃)²=( 9*10⁹*3*10⁻⁹*5*10⁻⁹) /(0.05) ²

F₁₃ = 5.4 *10⁻⁵ N

F₂₃ = (k*q₂*q₃)/(d₂₃)² = ( 9*10⁹*2*10⁻⁹*5*10⁻⁹) /( 0.03)²

F₂₃ = 10*10⁻⁵ N

x-y components of F₁₃ and F₂₃

F₁₃x= -5.4 *10⁻⁵ *cos β= - 5.4 *10⁻⁵ *(3/5)= - 3.24 *10⁻⁵ N

F₁₃y= -5.4 *10⁻⁵ *sin β= - 5.4 *10⁻⁵ *(4/5) = - 4.3 *10⁻⁵ N

F₂₃x = F₂₃ = + 10 *10⁻⁵ N

F₂₃y = 0

x and y components of the total force exerted on q₃ by q₁ and q₂ (Fn)

Fnx= F₁₃x+F₂₃x = - 3.24*10⁻⁵ N+10 *10⁻⁵ N= 6.76*10⁻⁵ N

Fny= F₁₃y+F₂₃y = - 4.3*10⁻⁵ N+0= - 4.3*10⁻⁵ N

Fn magnitude

`F_(n) = \sqrt{(Fn_(x))^(2) +(Fn_(y))^(2) }`

`F_(n) = \sqrt{(6.76)^(2)+(4.3)^(2)  } *10^(-5) N`

Fn= 8*10⁻⁵ N

Fn direction (α)

`\alpha = tan^(-1) ((Fn_(y) )/(Fn_(x) ) )`

`\alpha = tan^(-1) ((-4.3)/(6.76) )`

α = -32.46°

α = 32.46°, With respect to the positive axis of the x, in the fourth quadrant x-y