JEE Main & Advanced
JEE Main Paper (Held On 10 April 2016)
question_answer
Within a spherical charge distribution of charge density\[\rho (r)\], N equipotential surfaces of potential\[{{V}_{0}},\,{{V}_{0}}+\Delta V,{{V}_{0}}+2\Delta v\], ???..\[{{V}_{0}}+N\Delta V(\Delta V<0)\]are drawn and have increasing radii \[{{r}_{0}},{{r}_{1}}{{r}_{2}},.......{{r}_{N}}\],respectively. If the difference in the radii of the surface is constant for all values of \[{{V}_{0}}\] and \[\Delta V\] then:
JEE Main Online Paper (Held On 10 April 2016)
A)\[\rho (r)\alpha r\]
B)\[\rho (r)\alpha \frac{1}{{{r}^{2}}}\]
C)\[\rho (r)\alpha \frac{1}{r}\]
D)\[\rho (r)\]= constant
Correct Answer:
C
Solution :
\[\frac{\Delta V}{\Delta r}\to \]constant \[\Rightarrow \]uniform E. field. \[(E)\,(4\pi {{r}^{2}})=\frac{1}{{{\varepsilon }_{0}}}\int{\rho dV}\] \[(E)\,(4\pi {{r}^{2}})=\frac{1}{{{\varepsilon }_{0}}}\int\limits_{0}^{r}{\rho 4\pi {{r}^{2}}dr}\] \[(E)\,(4\pi {{r}^{2}})=\frac{1}{{{\varepsilon }_{0}}}4\pi \int\limits_{0}^{r}{\rho {{r}^{2}}dr}\] after integral on RHS We must obtain \[{{r}^{2}}\] \[\Rightarrow \,\rho \,\propto \frac{1}{r}\]