Coulomb's Law & Gradient Function: Can Anyone Explain?

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Coulomb's Law describes the electrostatic force between charged particles, stating that the force is proportional to the product of their charges and inversely proportional to the square of the distance between them. The gradient function measures the rate of change of a function, specifically applied here to 1/|x-x'|, where x and x' are particle positions. By rewriting 1/|x-x'| as |x-x'|^-1 and applying the power rule for derivatives, the gradient is derived as -(x-x')/|x-x'|^3. This transformation illustrates how the gradient function relates to Coulomb's Law, as it reflects changes in electric force with varying distances. Understanding this relationship is crucial for analyzing electric forces in physics.
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I was reading this ebook when I found Coulomb's law:
http://www.brokendream.net/xh4/elec.jpg
What I'm unsure of is how the gradient function applied to 1/|x-x'| is the same thing as |x-x'|/|x-x'|^3...
I know what the gradient function is but I'm totally lost as to how the formula was transformed that way..
Can anyone care to explain?
Thanks alot.
 
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Compute

\nabla_{\vec{r}}\left(\frac{1}{r}\right)

,for

r=:\sqrt{(x-x')^{2}+(y-y')^{2}+(z-z')^{2}}

and

\nabla_{\vec{r}}=\frac{\partial}{\partial x} \vec{i}+ \frac{\partial}{\partial y} \vec{j} +\frac{\partial}{\partial z} \vec{k}

Daniel.
 
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Coulomb's Law and the gradient function are both fundamental concepts in physics that are used to understand and describe electric forces. Coulomb's Law states that the magnitude of the electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This is represented by the equation F = kQ1Q2/r^2, where F is the force, k is a constant, Q1 and Q2 are the charges of the particles, and r is the distance between them.

The gradient function, on the other hand, is a mathematical tool used to measure the rate of change of a function at a particular point. In this case, we are looking at the function 1/|x-x'|, where x and x' represent the positions of two charged particles. The gradient function of this function is represented as ∇(1/|x-x'|).

To understand how this relates to Coulomb's Law, we can rewrite the function 1/|x-x'| as |x-x'|^-1. Then, using the power rule for derivatives, we can find the gradient function as ∇(|x-x'|^-1) = -|x-x'|^-2∇(|x-x'|) = -|x-x'|^-2(x-x')/|x-x'| = -(x-x')/|x-x'|^3.

As you can see, this is similar to the expression in Coulomb's Law, where the distance between the particles is represented as r = |x-x'|. So, the gradient function is essentially measuring the rate of change of the electric force between two charged particles as their distance changes.

I hope this explanation helps to clarify the relationship between Coulomb's Law and the gradient function. Both are essential in understanding and analyzing electric forces in physics.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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