Question

hiiiiii guys....... how to find out gravitational potential due to spherical shell??.... spam will be reported​

Answer 1

Answer:

Let radius of shell be R .

Then at a distance of r > R :

$\int \: dV = \int \: - E \: dr$

Putting the limits (coming from infinity to r )

$\displaystyle \: \int_{0}^{V} \: dV = \int_{\infty}^{r} \: - E \: dr$

$\displaystyle \: = > \int_{0}^{V} \: dV = \int_{\infty}^{r} \: - \frac{kq}{ {r}^{2} } \: dr$

$= > V = \dfrac{kq}{r} .......(r > R)$

Now for r = R , we can say :

$= > V = \dfrac{kq}{R} .......(r = R)$

For r < R , we know that inside a spherical shell, there is no Electric Field Intensity.

$\displaystyle \: \int_{V_{c}}^{V} \: dV = \int_{R}^{r} \: - E \: dr$

$\displaystyle \: = > \int_{V_{c}}^{V} \: dV = \int_{R}^{r} \: - (0) \: dr$

$= > V = V_{c}$

$= > V = \dfrac{kq}{R} .......(r < R)$

Answer 2

$</p><p>\huge{\tt{\pink{Hello!!!}}}$

$</p><p>\tt{\red{\underline {Finding \: Gravitational \: Potential \: Due \: To \: Spherical \: Shell \: - }}}$

$</p><p>\tt{\purple{Newton's \: Shell \: Theorem }}$

$</p><p>\tt{\green{Let\: the \: radius\: of \: the\:shell \:be\: R . }}$

$</p><p>\tt{\purple{Then \:at\: a \:distance\: of\: r \:>\: R \: - }}$

$</p><p>\orange{\mathfrak {\int \: dV = \int \: - E \: dr∫dV=∫_Edr}}$

Putting the limits (coming from infinity to r ).......

$</p><p>\red{\mathfrak {\displaystyle \: \int_{0}^{V} \: dV = \int_{\infty}^{r} \: -}}$

$</p><p>\purple{\mathfrak {\displaystyle \: = > \int_{0}^{V} \: dV = \int_{\infty}^{r} \: - \frac{kq}{ {r}^{2} } \: dr=>∫ }}$

$</p><p>\tt{\orange {= > V = \dfrac{kq}{r} .......(r > R)}}$

• Now for r = R :

$</p><p>\tt{\green{= > V = \dfrac{kq}{R} .......(r = R)}}$

For r < R , there is no Electric Field Intensity Inside A Spherical Shell....

$</p><p>\orange {\mathfrak {\displaystyle \: \int_{V_{c}}^{V} \: dV = \int_{R}^{r} \: - E \: dr∫ }}$

$</p><p>\tt{\blue {\displaystyle \: = > \int_{V_{c}}^{V} \: dV = \int_{R}^{r} \: - (0) \: dr=>∫ }}$

$</p><p>\tt{\purple{= > V = V_{c}=>V=V }}$

$</p><p>\tt{\red{= > V = \dfrac{kq}{R} .......(r < R)}}$