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JEE Advanced 2009Paper 1
Exam
JEE Advanced
Year
2009
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2009 · Paper 1 · Physics · Electrostatics

A disk of radius (a)/(4) having a uniformly distributed charge 6 C...

A disk of radius a4\frac{a}{4} having a uniformly distributed charge 6C6 \mathrm{C} is placed in the x−yx-y plane with its centre at \left(\frac{-a}{2}, 0,0\right).Arodoflengthacarryingauniformlydistributedcharge. A rod of length a carrying a uniformly distributed charge8 \mathrm{C}isplacedontheis placed on thex−axisfrom-axis fromx=\frac{a}{4} to x=\frac{5 a}{4}.Twopointcharges. Two point charges-7 \mathrm{C}andand3 \mathrm{C}areplacedatare placed at\left(\frac{a}{4}, \frac{-a}{4}, 0\right)and(−3a4,3a4,0)and \left(\frac{-3 a}{4}, \frac{3 a}{4}, 0\right), respectively. Consider a cubical surface formed by six surfaces x=±a2,y=±a2,z=±a2x=\pm \frac{a}{2}, y=\pm \frac{a}{2}, z=\pm \frac{a}{2}. The electric flux through this cubical surface is A disk of radius (a)/(4) having a uniformly distributed charge 6 C is placed in the x-y plane with its centre at

Options

  1. A−2Cε0\frac{-2 C}{\varepsilon_0}
  2. B2Cε0\frac{2 C}{\varepsilon_0}
  3. C10Cε0\frac{10 C}{\varepsilon_0}
  4. D12Cε0\frac{12 C}{\varepsilon_0}

Correct answer

A. −2Cε0\frac{-2 C}{\varepsilon_0}

Step-by-step solution

Total enclosed charge as already shown is q_{net \text{net }}=6C2+8C4−\frac{6 C}{2}+\frac{8 C}{4}-7 C=-2 C From Gauss-theorem, net flux, \varphi_{net \text{net }}=qnet ε0=−2Cε0\frac{q_{\text{net }}}{\varepsilon_0}=\frac{-2 C}{\varepsilon_0}
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