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JEE Main 2012 (AIEEE) · Electric Flux & Gauss's Law

Electric Flux & Gauss's Law question from JEE Main 2012 (AIEEE)

Q119JEE Main 2012 · AIEEEMedium

Electric Flux & Gauss's Law

This question has statement- 11 and statement- 2.2. Of the four choices given after the statements, choose the one that best describes the two statements.

An insulating solid sphere of radius RR has a uniformly positive charge density ρ\rho. As a result of this uniform charge distribution there is a finite value of electric potential at the center of the sphere, at the surface of the sphere and also at a point outside the sphere. The electric potential at infinity is zero.

Statement- 1:1: When a charge qq is taken from the centre to the surface of the sphere its potential energy changes by qρ3ε0\frac{q\rho}{3\varepsilon_0}

Statement- 2:2: The electric field at a distance r (r<R)r\,(r < R) from the center of the sphere is ρr3ε0\frac{\rho r}{3\varepsilon_0}

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Answer: (c)

  1. Statement 2: Gauss's law inside, E⋅4πr2=ρ43πr3ε0⇒E=ρr3ε0E \cdot 4\pi r^2 = \frac{\rho \frac{4}{3}\pi r^3}{\varepsilon_0} \Rightarrow E = \frac{\rho r}{3\varepsilon_0}. True.
  2. Statement 1: Vcentre−Vsurface=∫0Rρr3ε0dr=ρR26ε0V_{centre} - V_{surface} = \int_0^R \frac{\rho r}{3\varepsilon_0}dr = \frac{\rho R^2}{6\varepsilon_0}.
  3. So ΔU=qρR26ε0\Delta U = \frac{q\rho R^2}{6\varepsilon_0}, not qρ3ε0\frac{q\rho}{3\varepsilon_0} (that is not even in energy units). False.
  4. So the answer is (c) Statement 1 false, Statement 2 true.
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