Question

main potential energy of a particle in a field is U = a/r²-b/r, where a and b are constant. The value of r in terms of a and b where force on the particle is zero is equal to: xa/b find x?

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Answer 1

Answer:

• The value of x is 2

Explanation:

$\rule{300}{1.5}$

Given, we have a given equation of potential energy, and we need to evaluate it when the force is zero. So, from the relation of Potential energy and force we know that,

$\longrightarrow\sf F=-\Bigg[\dfrac{dU}{dx}\Bigg]$

Here,

• F Denotes Force.
• U Denotes Potential energy.

Solving,

$\longrightarrow\sf F=-\Bigg[\dfrac{d}{dx}\Bigg(\dfrac{a}{r^{2}}-\dfrac{b}{r}\Bigg)\Bigg]\\\\\\\\\longrightarrow\sf F=-\left[\dfrac{d\;\Big(ar^{-2}\Big)}{dx}-\dfrac{d\;\Big(br^{-1}\Big)}{dx}\right]$

• Differentiating it,

$\longrightarrow\sf F=-\Bigg[-2\times ar^{-3}-\bigg(-1\times br^{-2}\bigg)\Bigg]\\\\\\\\\longrightarrow\sf F=-\Bigg[\dfrac{-2a}{r^{3}}+\dfrac{b}{r^{2}}\Bigg]\\\\\\\\\longrightarrow\sf F=\Bigg[\dfrac{2a}{r^{3}}-\dfrac{b}{r^{2}}\Bigg]$

• Now, the Force on the particle is Zero.

$\longrightarrow\sf 0=\Bigg[\dfrac{2a}{r^{3}}-\dfrac{b}{r^{2}}\Bigg]\\\\\\\\\longrightarrow\sf \dfrac{2a}{r^{3}}=\dfrac{b}{r^{2}}\\\\\\\\\longrightarrow\sf \dfrac{2a}{r}=b\\\\\\\\\longrightarrow\sf r=\dfrac{2\;a}{b}$

• Comparing with r = xa/b

$\longrightarrow\sf \dfrac{x\;a}{b}=\dfrac{2\;a}{b}\\\\\\\\\longrightarrow\sf x=2\\\\\\\\\longrightarrow\large{\underline{\boxed{\red{\sf x=2}}}}$

Therefore, Value of x is 2.

$\rule{300}{1.5}$

Answer 2

Answer:

value of x is 2

Explanation:

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