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JEE Advanced 2019Paper 1
Exam
JEE Advanced
Year
2019
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2019 · Paper 1 · Physics · Electrostatics

A thin spherical insulating shell of radius R carries a uniformly...

A thin spherical insulating shell of radius RR carries a uniformly distributed charge such that the potential at its surface is V0{V}_{0} . A hole with a small area $\alpha4\pi{R}^{2}\left(\alpha ≪ 1\right) is made on the shell without affecting the rest of the shell. Which one of the following statements is correct?A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its

Options

  1. AThe magnitude of electric field at the center of the shell is reduced by αV02R\dfrac{\alpha{V}_{0}}{2R}
  2. BThe potential at the center of the shell is reduced by 2αV02\alpha{V}_{0}
  3. CThe ratio of the potential at the center of the shell to that of the point at 12R\dfrac{1}{2}R from center towards the hole will be 1−α1−2α\dfrac{1 - \alpha}{1 - 2\alpha}
  4. DThe magnitude of electric field at a point, located on a line passing through the hole and shell's center on a distance 2R2R from the center of the spherical shell will be reduced by αV02R\dfrac{\alpha{V}_{0}}{2R}

Correct answer

C. The ratio of the potential at the center of the shell to that of the point at 12R\dfrac{1}{2}R from center towards the hole will be 1−α1−2α\dfrac{1 - \alpha}{1 - 2\alpha}

Step-by-step solution

A thin spherical insulating shell of radius R carries a uniformly distributed charge such that the potential at its
Let ‘A’‘A’ be the point at a distance 2R2R from center and ‘B’‘B’ be the point at a distance of R2\dfrac{R}{2} from center.
Now, let total charge on spherical shell distributed uniformly =Q= Q
⇒Chargeperunitarea=Q4πr2⇒ Chargeperunitarea = \dfrac{Q}{4\pi{r}^{2}}
Charge on elemental area ‘dA’‘dA’ .
Given potential at surface =V0{= V}_{0}
⇒V0=KQR⇒ {V}_{0} = \dfrac{KQ}{R}
Now, potential at center of the shell =VC{= V}_{C}
VC=KQR−K(dQ)R=KQR−K(\alphaQ)R=KQR(1−α){V}_{C} = \dfrac{KQ}{R} - \dfrac{K ( dQ )}{R} = \dfrac{KQ}{R} - \dfrac{K ( \alphaQ )}{R} = \dfrac{KQ}{R} ( 1 - \alpha )
⇒VC=V0(1−α)⇒ReducedbyαV0⇒ {V}_{C} = {V}_{0}\left(1 - \alpha\right) ⇒ \text{Reduced}by\alpha{V}_{0}
Now potential =VB= {V}_{B}
⇒VB=KQR−K(aQ)R/2=V0(1−2a)⇒Reducedby2αV0⇒ {V}_{B} = \dfrac{KQ}{R} - \dfrac{K\left(aQ\right)}{R / 2} = {V}_{0}\left(1 - 2a\right) ⇒ Reducedby2\alpha{V}_{0}
∴VCVB=1−a1−2a∴ \dfrac{{V}_{C}}{{V}_{B}} = \dfrac{1 - a}{1 - 2a}
Now, electric field at A=EAA = {E}_{A}
Then, EA=KQ(2R)2−KaQR2=KQ4R2−aV0R{E}_{A} = \dfrac{KQ}{{\left(2R\right)}^{2}} - \dfrac{KaQ}{{R}^{2}} = \dfrac{KQ}{4{R}^{2}} - \dfrac{a{V}_{0}}{R}
So reduced by aV0R\dfrac{a{V}_{0}}{R}
Electric field at center ‘C’‘C’ due to small change ‘dQ’‘dQ’
EC=K(dq)R2=K(\alphaQ)R2=αV0R{E}_{C} = \dfrac{K\left(dq\right)}{{R}^{2}} = \dfrac{K\left(\alphaQ\right)}{{R}^{2}} = \dfrac{\alpha{V}_{0}}{R}
So increased by aV0R\dfrac{a{V}_{0}}{R}
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