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2. Electric potential and capacitance — Physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 SciencePhysics2. Electric potential and capacitance1 Mark
MCQ
Consider a gravity free container as shown. System is initially at rest and electric potential in the regon is $V = (y^3+2)\  J/C$. A ball of charge $q$ and mass $m$ is released from rest from base starts to move up due to electric field and collides with the shaded face as shown.If its speed just after collision is $1.5\  m/s$ and time for which ball is in contact with shaded face is $0.1\ sec$, find external force required to hold the container fixed in its position during collision assuming ball exerts constant force on wall during entire span of collision.......$N$
  • A
    $70$
  • B
    $72$
  • C
    $74$
  • ✓
    $76$
✓

Answer

Correct option: D.
$76$
d
$\mathrm{V}=\mathrm{y}^{3}+2$

$\Rightarrow \mathrm{E}=\frac{-\delta V}{\delta y} \hat{i}=-3 y^{2} \hat{j}$

$\mathrm{V}_{\mathrm{A}}=2 \mathrm{\,volt}$

$\mathrm{V}_{\mathrm{B}}=10$ volt $\left[\mathrm{V}=\mathrm{y}^{3}+2\right]$

$\mathrm{q}\left(\mathrm{V}_{\mathrm{B}}-\mathrm{V}_{\mathrm{A}}\right)=\frac{1}{2} m v^{2} \Rightarrow \frac{1}{2}(8)=\frac{1}{2}(2) \,V^{2}$

$\Rightarrow \mathrm{V}=2 \mathrm{\,m} / \mathrm{s}$

So, velocity of ball before collision $ = (2{\rm{\,m}}/{\rm{s}})j$

So, velocity of ball after collision $ =  - (1.5\,{\rm{m}}/{\rm{s}})j$

change in momentum $ = m\left( {{{\overrightarrow V }_F} - {{\overrightarrow V }_i}} \right) = ( - 7\,N.S)j$

Net force $ = ( - 7)/(0.1) = ( - 70\,{\rm{N}}){\rm{j}}$

from $\mathrm{FBD}$ of ball during collision

${{\rm{F}}_{{\rm{net}} = }}{\rm{ = }}{{\rm{F}}_{{\rm{wall}}}} - {\rm{qE}}$

$\mathrm{F}_{\mathrm{wall}}=\mathrm{F}_{\mathrm{net}}+\mathrm{q} \mathrm{E}$

$=(70+6)=76 \mathrm{\,N}$

${\rm{[E}}$ at top face $ = 3{{\rm{y}}^2} = 3{(2)^2} = 12\,{\mkern 1mu} {\rm{N}}/{\rm{C]}}$

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