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 There is a uniform spherically symmetric surface charge density at a distance Ro from the origin. The charge distribution is initially at rest and starts expanding because of mutual repulsion. The figure that represents best the speed V(R(t)) of the distribution as a function of its instantaneous radius R(t) is : 

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

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Potential Energy Of System Of two Charge -

U=\frac{kQ_{1}Q_{2}}{r}  \left ( S.I \right )

U= \frac{Q_{1}Q_{2}}{r}  \left ( C.G.S \right ) 

 

- wherein

K=\frac{1}{4\pi \epsilon _{0}}

At any instant

Total energy of charge distribution is constant

\frac{1}{2}mV^{2}+\frac{KQ^{2}}{2R}+0+\frac{KQ^{2}}{2R_{0}}

\frac{1}{2}mV^{2}+\frac{KQ^{2}}{2R_{0}}-\frac{KQ^{2}}{2R}

V=\sqrt{\frac{2}{m}\, \, \frac{KQ^{2}}{2}\left ( \frac{1}{R_{0}}-\frac{1}{R} \right )}

     =c\sqrt{\frac{1}{R_{0}}-\frac{1}{R}}

and slope of V-R curve is decreasing

 

 

 


Option 1)

Option 2)

Option 3)

Option 4)

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