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Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positivex=axis at x = a+r, a distance r to the right end of Q.(a) Calculate the x- and y-components of the electric fieldporduced by the charge distribution Q at points on the positivex-axis where x>a.(b) Calculate the force (magnitude and direction) that thecharge distribution Q exerts on q.(c) Show that if r>>a, the magnitude of the force inpart (b) is approximately Qq/(4pi epsilon0 r^2). Explain why thisresult is obtained.

Respuesta :

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Answer:

a. b- x= y

dx = -dy

b. F = [tex]\frac{-kQqi}{r (a+r)}[/tex]

c.  F = [tex]\frac{-kQqi}{r^{2} }[/tex]

Explanation:

a. x components:

[tex]dE = \frac{kdq}{(a+r-x^{2}) } \\[/tex]

     = [tex]\frac{kQdx}{(a(a+r-x)^2}[/tex]

Integrating and solving gives:

b- x= y

dx = -dy

b. the force is given by the equation derived from (a.):

F = [tex]\frac{-kQqi}{r (a+r)}[/tex]

c. Given that r>>a, the expression becomes:

F = [tex]\frac{-kQqi}{r^{2} }[/tex]

Explanation:

When the size of the charge distribution is less than the distance to the deviation point of the charge then the charge distribution would produce the same effect such as a linear charge.

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