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JEE Advanced 2010Paper 2
Exam
JEE Advanced
Year
2010
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2010 · Paper 2 · Physics · Electrostatics

A uniformly charged thin spherical shell of radius R carries...

A uniformly charged thin spherical shell of radius RR carries uniform surface charge density of σ\sigma per unit area. It is made of two hemispherical shells, held together by pressing them with force FF (see figure). FF is proportional to A uniformly charged thin spherical shell of radius R carries uniform surface charge density of σ per unit area. It is

Options

  1. A1ε0σ2R2\frac{1}{\varepsilon_0} \sigma^2 R^2
  2. B1ε0σ2R\frac{1}{\varepsilon_0} \sigma^2 R
  3. C1ε0σ2R\frac{1}{\varepsilon_0} \frac{\sigma^2}{R}
  4. D1ε0σ2R2\frac{1}{\varepsilon_0} \frac{\sigma^2}{R^2}

Correct answer

A. 1ε0σ2R2\frac{1}{\varepsilon_0} \sigma^2 R^2

Step-by-step solution

Electrical force per unit area =12ε0E2=\frac{1}{2} \varepsilon_0 E^2 =\frac{1}{2} \varepsilon_0$\left(\frac{\sigma}{\varepsilon_0}\right)^2=\frac{\sigma^2}{2 \varepsilon_0} Projected area =πR2∴=\pi R^2 \therefore net electrical force =\left(\frac{\sigma^2}{2 \varepsilon_0}\right)\left(\pi R^2\right)Inequilibrium,thisforceshouldbeequaltotheappliedforce.In equilibrium, this force should be equal to the applied force.\therefore \quad F=\frac{\pi \sigma^2 R^2}{2 \varepsilon_0}oror\quad F \propto \frac{\sigma^2 R^2}{\varepsilon_0}\therefore$ The correct option is (a).
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