Physics, 10.09.2019 19:30 queenhall934

There are four charges, each with a magnitude of 1.96 µc. two are positive and two are negative. the charges are fixed to the corners of a 0.47-m square, one to a corner, in such a way that the net force on any charge is directed toward the center of the square. find the magnitude of the net electrostatic force experienced by any charge.

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Answer from: cganderson04

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:

$F_{12} = \frac{k*q_1*q_2}{r_{12}^2}$ : Magnitude of the electrical force of q₂ over q₁. Equation((1)

$F_{13} = \frac{k*q_1*q_3}{r_{13}^2}$ : Magnitude of the electrical force of q₃ over q₁. Equation (2)

$F_{14} = \frac{k*q_1*q_4}{r_{14}^2}$ : 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

$r_{14} = \sqrt{0.47^2+ 0.47^2}=0.664m$

k=8.99x10⁹N*m²/C² : Coulomb constant

F₁₂ calculation

We replace data in the equation (1):

$F_{12} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}$

F₁₂ = 0.156 N Direction of the positive x axis (+x)

F₁₃ calculation

We replace data in the equation (2):

$F_{13} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.47^2}$

F₁₃ = 0.156 N Direction of the negative y axis (-y)

Magnitude of the net electrostatic force between F₁₃ and F₁₂

$F_{n23}= \sqrt{0.156^2+0.156^2} = 0.22N$ (directed toward the center of the square)

F₁₄ calculation

We replace F₁₄ data in the equation (3):

$F_{14} = \frac{8.99*10^9*(1.96*10^{-6})^2}{0.664^2}$

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

There are four charges, each with a magnitude of 1.96 µc. two are positive and two are negative. the

Answer from: Quest

It is the areas with lines. this may sound a little confusing, but i hope it !

Answer from: Quest

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