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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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Answer:

350N

Explanation:

Given parameters:

Mass of the man = 125kg

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Unknown:

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Solution:

To find this force we use;

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mass  = 125 + 6 + 3 + 6  = 140kg

 So;

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C is the awnser because c is single replacement
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marissa [1.9K]

Answer:

Newton's Third Law

Explanation:

Newton's third law

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6 0
2 years ago
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