Verified Solution • 2025–26 Session moderate

QJEEM19Phc433: JEE Main 2019 Physics - Class 12 JEEM19Phc433 Solution: There is a uniform spherically symmetric surface charge density …

There is a uniform spherically symmetric surface charge density at a distance R0 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 :

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/2bd643a219730ce6b6cf639da1315bc5202c2c9f457552089f8da1b77d74ffbb.webp
  • B https://ik.imagekit.io/novatra/photos/2c0ba5a44a237172de4b0829eb9630a06d9098aeb1388ec0073bb7bf879426c5.webp
  • C https://ik.imagekit.io/novatra/photos/d80d8ad54662c7ca136eef358d8f86feb56049bab755490520dfd451503b5376.webp
  • D https://ik.imagekit.io/novatra/photos/82065763b3cfc7cf0302a8fc21e4e6c80aa199547e79fd851e4b7d05a79e4ce3.webp

Step-by-Step Verified Solution

At any instant 't'

Total energy of charge distribution is constant

i.e.      12mV2+KQ22R=0+KQ22R0{1 \over 2}m{V^2} + {{K{Q^2}} \over {2R}} = 0 + {{K{Q^2}} \over {2{R_0}}}

∴\therefore    12mV2=KQ22R0−KQ22R{1 \over 2}m{V^2} = {{K{Q^2}} \over {2{R_0}}} - {{K{Q^2}} \over {2R}}

∴\therefore   V = 2mKQ22.(1R0−1R)\sqrt {{2 \over m}{{K{Q^2}} \over 2}.\left( {{1 \over {{R_0}}} - {1 \over R}} \right)}

∴\therefore   V = KQ2m(1R0−1R)\sqrt {{{K{Q^2}} \over m}\left( {{1 \over {{R_0}}} - {1 \over R}} \right)} = C 1R0−1R\sqrt {{1 \over {{R_0}}} - {1 \over R}}

Also the slope of v-s curve will go on decreasing

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