Answer:
Magnitude of the resultant force (Fn₁) on q₁
Fn₁ = 0.142N (directed toward the center of the square)
Explanation:
Theory of electrical forces
Because the particle q₁ is close to three other electrically charged particles, it will experience three electrical forces and the solution of the problem is of a vector nature.
Graphic attached
The directions of the individual forces exerted by q₂, q₃ and q₄ on q₁ are shown in the attached figure.
The force (F₁₄) of q₄ on q₁ is repulsive because the charges have equal signs and the forces (F₁₂) and (F₁₃) of q₂ and q₃ on q₁ are attractive because the charges have opposite signs.
Calculation of the forces exerted on the charge q₁
To calculate the magnitudes of the forces exerted by the charges q₂, q₃, and q₄ on the charge q₁ we apply Coulomb's law:
[tex]F_{12} = \frac{k*q_1*q_2}{r_{12}^2}[/tex]: Magnitude of the electrical force of q₂ over q₁. Equation((1)
[tex]F_{13} = \frac{k*q_1*q_3}{r_{13}^2}[/tex]: Magnitude of the electrical force of q₃ over q₁. Equation (2)
[tex]F_{14} = \frac{k*q_1*q_4}{r_{14}^2}[/tex]: Magnitude of the electrical force of q₄ over q₁. Equation (3)
Equivalences
1µC= 10⁻⁶ C
Known data
q₁=q₄= 1.96 µC = 1.96*10⁻⁶C
q₂=q₃= -1.96 µC = -1.96*10⁻⁶C
r₁₂= r₁₃ = 0.47m: distance between q₁ and q₂ and q₁ and q₄
[tex]r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m[/tex]
k=8.99x10⁹N*m²/C² : Coulomb constant
F₁₂ calculation
We replace data in the equation (1):
[tex]F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]
F₁₂ = 0.156 N Direction of the positive x axis (+x)
F₁₃ calculation
We replace data in the equation (2):
[tex]F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}[/tex]
F₁₃ = 0.156 N Direction of the negative y axis (-y)
Magnitude of the net electrostatic force between F₁₃ and F₁₂
[tex]F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N[/tex] (directed toward the center of the square)
F₁₄ calculation
We replace F₁₄ data in the equation (3):
[tex]F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}[/tex]
F₁₄ = 0.078 N (In the opposite direction to Fn₂₃)
Calculation of the resulting force on q₁: Fn₁
Fn₁ = Fn₂₃ - F₁₄ = 0.22 - 0.078 = 0.142 N
