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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Capacitance

JIPMER 2018 · Physics · Capacitance

Find the capacitance in shown figure

Find the capacitance in shown figure Find the capacitance in shown figure

Options

  1. A2KAε0(K+1)d\frac{2 K A \varepsilon_0}{(K+1) d}
  2. B2KAε0d\frac{2 K A \varepsilon_0}{d}
  3. C(K+1)Aε02d\frac{(K+1) A \varepsilon_0}{2 d}
  4. D2KAε0(K2+1)d\frac{2 K A \varepsilon_0}{\left(K^2+1\right) d}

Correct answer

A. 2KAε0(K+1)d\frac{2 K A \varepsilon_0}{(K+1) d}

Step-by-step solution

This combination is same as the two capacitors are connected in series and distance between plates of each capacitor is d/2\mathrm{d} / 2. So, C1=Kε0Ad/2\mathrm{C}_1=\frac{\mathrm{K} \varepsilon_0 \mathrm{ A}}{\mathrm{ d} / 2} and C2=ε0Ad/2\mathrm{C}_2=\frac{\varepsilon_0 \mathrm{ A}}{\mathrm{ d} / 2} Hence, Cnet =C1C2C1+C2=(2Kε0Ad)(2ε0Ad)(2Kε0Ad+2ε0Ad)=2KAε0(K+l)dC_{\text{net }}=\frac{C_1 C_2}{C_1+C_2}=\frac{\left(\frac{2 K \varepsilon_0 \mathrm{ A}}{d}\right)\left(\frac{2 \varepsilon_0 \mathrm{ A}}{d}\right)}{\left(\frac{2 K \varepsilon_0 \mathrm{ A}}{d}+\frac{2 \varepsilon_0 \mathrm{ A}}{d}\right)} =\frac{2 \mathrm{KA} \varepsilon_0}{(\mathrm{ K}+\mathrm{l}) \mathrm{d}}
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