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Fudgin [204]
3 years ago
6

A charge Q is to be divided into two parts, labeled 'q' and 'Q-q? What is the relationship of q to Q if, at any given distance,

the Coulomb force between the two is to be maximized?
Physics
1 answer:
Lisa [10]3 years ago
6 0

Answer:

Explanation:

The two charges are q and Q - q. Let the distance between them is r

Use the formula for coulomb's law for the force between the two charges

F = \frac{Kq_{1}q_{2}}{r^{2}}

So, the force between the charges q and Q - q is given by

F = \frac{K\left ( Q-q \right )q}}{r^{2}}

For maxima and minima, differentiate the force with respect to q.

\frac{dF}{dq} = \frac{k}{r^{2}}\times \left ( Q - 2q \right )

For maxima and minima, the value of dF/dq = 0

So, we get

q = Q /2

Now \frac{d^{2}F}{dq^{2}} = \frac{-2k}{r^{2}}

the double derivate is negative, so the force is maxima when q = Q / 2 .

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Three beads are placed along a thin rod. The first bead, of mass m1 = 23 g, is placed a distance d1 = 1.1 cm from the left end o
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Answer:

a) x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

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c) x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

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Explanation:

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x=\frac{m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3} }{m_{1}+m_{2} +m_{3}}

Where m is the mass of beads and x is the distances, if x₁ = d₁, x₂ = d₂ and x₃ = d₃

x=\frac{m_{1}d_{1}+m_{2}(d_{1}+d_{2})+m_{3}(d_{1}+d_{2}+d_{3}  ) }{m_{1}+m_{2}+m_{3} }

b) If

m₁ = 23g

m₂ = 15 g

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d₃ = 3.2 cm

x=\frac{23*1.1+15*(1.1+1.9)+58(1.1+1.9+3.2) }{23+15+58 } =4.47cm

c) The center of the mass of the beads realtive to the center of bead is:

x=\frac{m_{1}d_{2}+m_{2}(0)+m_{3}d_{3} }{m_{1}+m_{2}+m_{3} }

d) x=\frac{23*(-1.9)+(15*0)+(58*3.2) }{23+15+58 } =1.48cm

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Answer:

d)

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