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JIPMER 2009
Exam
JIPMER
Year
2009
Subject
Physics
Chapter
Electrostatics

JIPMER 2009 · Physics · Electrostatics

Two spherical conductors B and C having equal radii and carrying...

Two spherical conductors BB and CC having equal radii and carrying equal charges in them repel each other with a force FF when kept apart at some distance. A third spherical conductor having same radius as that of BB but uncharged, is brought in contact with BB, then brought in contact with CC and finally removed away from both. The new force of repulsion between BB and CC is

Options

  1. AF4\frac{F}{4}
  2. B3F4\frac{3 F}{4}
  3. CF8\frac{F}{8}
  4. D3F8\frac{3 F}{8}

Correct answer

D. 3F8\frac{3 F}{8}

Step-by-step solution

Let the spherical conductors BandCB and C have same charge as qq. The electric force between them is F=14πε0q2r2rF=\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2} r, being the distance between them. When third uncharged conductor AA is brought in contact with BB, then charge on each conductor qA=qB=qA+qB2=0+q2=q2q_A=q_B=\frac{q_A+q_B}{2}=\frac{0+q}{2}=\frac{q}{2} When this conductor AA is now brought in contact with CC, then charge on each conductor qA=qC=qA+qC2=(q/2)+q2=3q4q_A=q_C=\frac{q_A+q_C}{2}=\frac{(q / 2)+q}{2}=\frac{3 q}{4} Hence, electric force acting between BandCB and C is F′=14πε0qBqCr2=14πε0(q/2)(3q/4)r2=38[14πε0q2r2]=3F8\begin{aligned} F^{\prime}=\frac{1}{4 \pi \varepsilon_0} \frac{q_B q_C}{r^2} & =\frac{1}{4 \pi \varepsilon_0} \frac{(q / 2)(3 q / 4)}{r^2} \\ & =\frac{3}{8}\left[\frac{1}{4 \pi \varepsilon_0} \frac{q^2}{r^2}\right]=\frac{3 F}{8} \end{aligned}
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