question_answer

Three concentric spherical shells have radii a, b and c( a < b < c) and have surface charge densities   \[\therefore \]and \[y=2\sin (7.85\times -1005t)\] respectively.   If \[=(0.02)m\sin (7.85x-1005t)\] and \[v=\frac{dy}{dt}=A\omega \cos \omega t=A\omega \sqrt{1-{{\sin }^{2}}\omega t}\]denote the potentials of the three shells, then for c = a + b, we have

A) \[=\omega \sqrt{{{A}^{2}}-{{y}^{2}}}\]  

B)        \[y=\frac{a}{2}\]

C) \[\therefore \]                 

D) \[v=\omega \sqrt{{{a}^{2}}-\frac{{{a}^{2}}}{4}}=\omega \sqrt{\frac{3{{a}^{2}}}{4}}=\frac{2\pi }{T}\frac{a\sqrt{3}}{2}=\frac{\pi a\sqrt{3}}{T}\]

Correct Answer: A

Solution :

Hence, \[\to \] \[\to \] \[\to \]\[\to \] \[{{V}_{B}}=\frac{1}{4\pi {{\varepsilon }_{0}}}\,-\,\frac{\sigma 4\pi {{\alpha }^{2}}}{a}\,-\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{\sigma 4\pi {{b}^{2}}}{b}\] \[\to \] \[=\frac{\sigma }{{{\varepsilon }_{0}}}\,\left( \frac{{{a}^{2}}}{c}-b+c \right)\,=\frac{\sigma }{{{\varepsilon }_{0}}}(2a)\,\,(\because \,c=a+b)\]and\[\to \]\[\to \] \[\to \] Hence,\[\to \]