R-x-2rcfrac - Gmm-x-r- - - r - 3 - - cfrac - Gmm-x-r- - 2- r - 3 - - cfrac - Gmm-2x-r- - - r - 3 - - cfrac - Gmm-2r-x- - 2- r - 3 - -

Question

Toppr Admin · Accepted Answer

Since particle is inside the shell- there is no force exerted by shell on the particle-Gravitational force due to sphere-Gmenclosedm-x2032-x-x2212-r-2-Gm-x-x2212-r-3r3m-x2032-x-x2212-r-2-Gmm-x2032-x-x2212-r-r3

Table 1	Table - 2
(a) current through $R_{1}$	(p) E/R
(b) current through $R_{2}$	(q) 2E/R
(c) current through $R_{3}$	(r) E/2R
	(s) Zero

$r < x < 2 r$

$\frac{G m m^{'} (x - r)}{r^{3}}$
$\frac{G m m^{'} (x - r)}{2 r^{3}}$
$\frac{G m m^{'} (2 x - r)}{r^{3}}$
$\frac{G m m^{'} (2 r - x)}{2 r^{3}}$

Since particle is inside the shell, there is no force exerted by shell on the particle.
Gravitational force due to sphere= $\frac{G m_{e n c l o s e d} m^{'}}{(x - r)^{2}}$
= $\frac{G m \frac{(x - r)^{3}}{r^{3}} m^{'}}{(x - r)^{2}}$
$= \frac{G m m^{'} (x - r)}{r^{3}}$

r<x<2rGmm′(x−r)r3Gmm′(x−r)2r3Gmm′(2x−r)r3Gmm′(2r−x)2r3

Since particle is inside the shell, there is no force exerted by shell on the particle.Gravitational force due to sphere=Gmenclosedm′(x−r)2=Gm(x−r)3r3m′(x−r)2=Gmm′(x−r)r3

$r < x < 2 r$

$\frac{G m m^{'} (x - r)}{r^{3}}$
$\frac{G m m^{'} (x - r)}{2 r^{3}}$
$\frac{G m m^{'} (2 x - r)}{r^{3}}$
$\frac{G m m^{'} (2 r - x)}{2 r^{3}}$

Since particle is inside the shell, there is no force exerted by shell on the particle.
Gravitational force due to sphere= $\frac{G m_{e n c l o s e d} m^{'}}{(x - r)^{2}}$
= $\frac{G m \frac{(x - r)^{3}}{r^{3}} m^{'}}{(x - r)^{2}}$
$= \frac{G m m^{'} (x - r)}{r^{3}}$