Answer:
(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
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
[tex]d_{13} =\sqrt{3^{2}+4^{2} } = 5cm = 0.05m[/tex]
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
[tex]F_{n} = \sqrt{(Fn_{x})^{2} +(Fn_{y})^{2} }[/tex]
[tex]F_{n} = \sqrt{(6.76)^{2}+(4.3)^{2} } *10^{-5} N[/tex]
Fn= 8*10⁻⁵ N
Fn direction (α)
[tex]\alpha = tan^{-1} (\frac{Fn_{y} }{Fn_{x} } )[/tex]
[tex]\alpha = tan^{-1} (\frac{-4.3}{6.76} )[/tex]
α = -32.46°
α = 32.46°, With respect to the positive axis of the x, in the fourth quadrant x-y
