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JEE Advanced 2008Paper 2
Exam
JEE Advanced
Year
2008
Paper
Paper 2
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2008 · Paper 2 · Physics · Electrostatics

Consider a system of three charges (q)/(3), (q)/(3) and -(2 q)/(3)...

Consider a system of three charges q3,q3\frac{q}{3}, \frac{q}{3} and −2q3-\frac{2 q}{3} placed at points A,BA, B and CC, respectively, as shown in the figure. Take OO to be the centre of the circle of radius RR and angle CAB=60∘C A B=60^{\circ}. Consider a system of three charges (q)/(3), (q)/(3) and -(2 q)/(3) placed at points A, B and C , respectively, as shown

Options

  1. AThe electric field at point OO is q8πε0R2\frac{q}{8 \pi \varepsilon_0 R^2} directed along the negative xx-axis
  2. BThe potential energy of the system is zero
  3. CThe magnitude of the force between the charges at CC and BB is q254πε0R2\frac{q^2}{54 \pi \varepsilon_0 R^2}
  4. DThe potential at point OO is q12πε0R\frac{q}{12 \pi \varepsilon_0 R}

Correct answer

C. The magnitude of the force between the charges at CC and BB is q254πε0R2\frac{q^2}{54 \pi \varepsilon_0 R^2}

Step-by-step solution

Distance BC=ABsin⁡60∘=(2R)32=3RB C=A B \sin 60^{\circ}=(2 R) \frac{\sqrt{3}}{2}=\sqrt{3} R ∴∣FBC∣=14πε0(q/3)(2q/3)(3R)2=q254πε0R2\therefore \quad\left|F_{B C}\right|=\frac{1}{4 \pi \varepsilon_0} \frac{(q / 3)(2 q / 3)}{(\sqrt{3} R)^2}=\frac{q^2}{54 \pi \varepsilon_0 R^2} ∴\therefore correct option is (c).
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