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In Fig. 21-38, particle 1 of charge +4eis above a floor by distance d1=2.00mmand particle 2 of charge+6eis on the floor, at distanced2=6.00mmhorizontally from particle 1.What is the xcomponent of the electrostatic force on particle 2 due to particle 1?

Short Answer

Expert verified

The x-component of the electrostatic force on particle 2 due to particle 1 is1.31×10−22 N

Step by step solution

01

The given data

  1. Particle 1 of charge+4e is above a floor by distance,d1=2″¾³¾´Ç°ù0.002″¾
  2. Particle 2 of charge+6e is on the floor, at distanced2=6″¾³¾´Ç°ù0.006″¾ horizontally from particle 1.
02

Understanding the concept of Coulomb’s law 

Using the given charges in the formula of force from Coulomb's law, we can get the x-component of force on particle 2 due to particle 1. To get the x-component of the force, we can get the unit-vector of the distance.

Formulae:

The magnitude of the electrostatic force between any two particles,

F21=|q1q2|4πε0r2,wherer=d12+d22 (i)

The distance vector in unit-vector notation, r^=rr→=d2i^−d1j^d12+d22 (ii)

03

Calculation of the x-component of the electrostatic force            

Since bothq1andq2are positively charged, particle 2 is repelled by particle 1, so the direction of F21is away from particle 1 and toward 2.

In unit-vector notation, the force acting on the particle 2 due to 1 can be given as:F12→=F12r^

Now, for the x-component of the force, we can take the i^vector value of the distance from equation (ii) as given:

F→12(x)=F12d2d12+d22=kq1q2d2(d12+d22)3/2(substitutingequation(i))=(9×109Nâ‹…m2C2)(4×1.60×10−19 C)(6×1.60×10−19 C)(6.00×10−3″¾)[(2.00×10−3″¾)2+(6.00×10−3″¾)2]32=1.31×10−22 N

Hence, the x-component of the force is1.31×10−22 N

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