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OverLord2011 [107]
3 years ago
6

Positive charge Q is distributed uniformly along the x-axis fromx=0 to x=a. A positive point charge q is located on the positive

x=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.
Physics
1 answer:
dybincka [34]3 years ago
6 0

Answer:

a. b- x= y

dx = -dy

b. F = \frac{-kQqi}{r (a+r)}

c.  F = \frac{-kQqi}{r^{2} }

Explanation:

a. x components:

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

     = \frac{kQdx}{(a(a+r-x)^2}

Integrating and solving gives:

b- x= y

dx = -dy

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

F = \frac{-kQqi}{r (a+r)}

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

F = \frac{-kQqi}{r^{2} }

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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Answer:

5.01 J

Explanation:

Info given:

mass (m) = 0.0780kg

height (h) = 5.36m

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1. First, solve for Kinetic energy (KE)

KE = 1/2mv^2

1/2(0.0780kg)(4.84m/s)^2 = 0.91 J

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4 0
3 years ago
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lawyer [7]

Answer:

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8 0
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Answer:

Explanation:

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