Question

Gravitational field intensity at point P which is at a distance x from the centre of the ring is

Answer 1

Field Intensity in axis of ring:

• Consider a charge dm on the ring exerting dE field at the axis. Their will be a diametrically opposite dm that will also exert dE.

• Now, the sine components will be cancelled.

$dE_{net} = dE \cos( \theta)$

$\implies dE_{net} = \dfrac{G(dm)}{ {r}^{2} + {x}^{2} } \times \dfrac{x}{ \sqrt{ {r}^{2} + {x}^{2} }}$

$\displaystyle \implies \int dE_{net} = \int\dfrac{G(dm)}{ {r}^{2} + {x}^{2} } \times \dfrac{x}{ \sqrt{ {r}^{2} + {x}^{2} }}$

$\displaystyle \implies \int dE_{net} = \dfrac{Gx}{ {({r}^{2} + {x}^{2})}^{ \frac{3}{2} } } \: \int \: dm$

$\boxed{ \displaystyle \implies E_{net} = \dfrac{Gmx}{ {({r}^{2} + {x}^{2})}^{ \frac{3}{2} } } }$

Hope It Helps.

Answer 2

Answer:

Explanation:

Field Intensity in axis of ring:

Consider a charge dm on the ring exerting dE field at the axis. Their will be a diametrically opposite dm that will also exert dE.

Now, the sine components will be cancelled.