Field at a distance x due to a circular ring- EG - -GM-R2 - x23-

Let the linear mass velocity of the ring be #160; λ Gravitational potential #160; = GM √( r ^ 2 + x ^ 2 ) #160; As potential is scalar quantity it ...

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Standard XII

Physics

Gravitational Potential

Field at a di...

Question

Field at a distance $x$ due to a circular ring, $E_{G} = \frac{- G M}{\sqrt{(R^{2} + x^{2})^{3}}}$ .

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Solution

Let the linear mass velocity of the ring be $λ$
Gravitational potential $= \frac{G M}{\sqrt{r^{2} + x^{2}}}$
As potential is scalar quantity it will be added for each unit mass.
Now, $E = \frac{- d V}{d x} = \frac{- d (\frac{G M}{\sqrt{r^{2} + x^{2}}})}{d x}$
$E = \frac{- G M}{\sqrt{{(R^{2} + x^{2})}^{3}}}$

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Field at a distance x due to a circular ring, EG=−GM√(R2+x2)3.

Let the linear mass velocity of the ring be λGravitational potential =GM√r2+x2As potential is scalar quantity it will be added for each unit mass.Now, E=−dVdx=−d(GM√r2+x2)dxE=−GM√(R2+x2)3

Field at a distance $x$ due to a circular ring, $E_{G} = \frac{- G M}{\sqrt{(R^{2} + x^{2})^{3}}}$ .