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Expression of the force derived from this potential
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[QUOTE="Delta2, post: 6412063, member: 189563"] It is a well known fact that given the potential function ##U(r,\theta,\phi)## of a field , the force that the field applies to a test point charge (or test point mass or whatever is the field subject)## q## is given by $$\vec{F}=-q\nabla U$$. So all you have to do is calculate the gradient ##\nabla U## of the function ##U##. I assume from the context that the calculation must be done in spherical coordinates. So it will be $$\nabla U=\frac{\partial U}{\partial r}\hat r+\frac{1}{r}\frac{\partial U}{\partial \theta}\hat\theta+\frac{1}{r\sin\theta}\frac{\partial U}{\partial \phi}\hat\phi$$ Also note that in your case the function U does not depend on ##\theta## and ##\phi## so the above formula for the gradient of U simplifies a lot. [/QUOTE]
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Expression of the force derived from this potential
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