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What is (Fnet3)x , the x-component of the net force exerted by these two charges on a third charge q3 = 50.5 nC placed between q1 and q2 at x3 = -1.190 m ?

Your answer may be positive or negative, depending on the direction of the force.

What Is Fnet3x The Xcomponent Of The Net Force Exerted By These Two Charges On A Third Charge Q3 505 NC Placed Between Q1 And Q2 At X3 1190 M Your Answer May Be class=

Sagot :

whitneytr12

The x-component of the net force exerted by the first (q₁ = -11.0 nC) and second charge (q₂ = 36.0 nC) on a third one (q₃ = 50.5 nC), located between q₁ (x₁ = -1.740 m) and q₂ (x₂ = 0 m) at x₃ = -1.190 m is -5.00x10⁻⁶ N.    

The x-component of the net force exerted by the first and second charge on the third charge is given by:

[tex] \vec{F}_{net_{x}} = \vec{F}_{13} + \vec{F}_{23} [/tex]   (1)  

Where:

  • [tex]\vec{F}_{13}[/tex]: is the force exerted by the first charge on the third.
  • [tex] \vec{F}_{23}[/tex]: is the force exerted by the second charge on the third.  

Calculation of the force exerted by each charge on the third one

1. Force exerted by the first charge on the third

We can calculate the force with Coulomb's law

[tex] \vec{F}_{13} = \frac{Kq_{1}q_{3}}{d_{13}^{2}} [/tex] (2)  

Where:

  • q₁: is the first charge = -11.0x10⁻⁹ C
  • q₃: is the third charge = 50.5x10⁻⁹ C
  • d₁₃: is the distance between charges 1 and 3
  • K: is the Coulomb's constant

The Coulomb's constant is:

[tex] K = \frac{1}{4\pi \epsilon_{0}} = \frac{1}{4\pi 8.854\cdot 10^{-12} C^{2}/N*m^{2}} = 8.99 \cdot 10^{9} N*m^{2}/C^{2} [/tex]    

Since the third charge is placed between q₁ and q₂ at x₃ = -1.190 m, the magnitude of the distance d₁₃, knowing that x₁ = -1.740 m, is:

[tex] d_{13} = |d_{1}| - |d_{3}| = (1.740 - 1.190) m = 0.55 m [/tex]

Hence, the force exerted by the first charge on the third is (eq 2):

[tex] \vec{F}_{13} = \frac{Kq_{1}q_{3}}{d_{13}^{2}} = \frac{8.99 \cdot 10^{9} N*m^{2}/C^{2}*(-11.0 \cdot 10^{-9} C)(50.5 \cdot 10^{-9} C)}{(0.55 m)^{2}} = -1.65 \cdot 10^{-5} N [/tex]  

The minus sign is because the resultant force is in the negative x-direction (to the left).

2. Force exerted by the second charge on the third

This force is equal to:

[tex] \vec{F}_{23} = \frac{Kq_{2}q_{3}}{d_{23}^{2}} [/tex]   (3)

Where:

  • q₂: is the second charge = 36.0x10⁻⁹ C
  • d₂₃: is the distance between charges 2 and 3

The magnitude of the distance d₂₃, knowing that x₂ = 0, is:

[tex] d_{23} = |d_{3}| - |d_{2}| = (1.190 - 0) m = 1.190 m [/tex]  

So, the force exerted by the second charge on the third is (eq 3):

[tex] \vec{F}_{23} = \frac{Kq_{2}q_{3}}{d_{23}^{2}} = \frac{8.99 \cdot 10^{9} N*m^{2}/C^{2}*36.0 \cdot 10^{-9} C*50.5 \cdot 10^{-9} C}{(1.190 m)^{2}} = 1.15 \cdot 10^{-5} N [/tex]    

The positive sign means that the resultant force is in the positive x-direction (to the right).  

Finally, the x-component of the net force is (eq 1):

[tex] \vec{F}_{net_{x}} = \vec{F}_{13} + \vec{F}_{23} = (-1.65 \cdot 10^{-5} + 1.15 \cdot 10^{-5}) N = -5.00 \cdot 10^{-6} N [/tex]

The minus sign means that the net force exerted by these two charges on the third charge is in the negative x-direction.  

Therefore, the x-component of the net force exerted by the first and second charge on a third charge is -5.00x10⁻⁶ N.

Find more about Coulomb's law here:

brainly.com/question/506926

I hope it helps you!

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