The capacitance of a parallel plate capacitor is $\mathrm{C}_{\mathrm{a}}$ (Fig. a). A dielectric of…

The capacitance of a parallel plate capacitor is $\mathrm{C}_{\mathrm{a}}$ (Fig. a). A dielectric of dielectric constant $\mathrm{K}$ is inserted as shown in fig (b) and (c) then
  1. both \(\mathrm{C}_{\mathrm{b}}, \mathrm{C}_{\mathrm{c}} > \mathrm{C}_{\mathrm{a}}\)
  2. \(\mathrm{C}_{\mathrm{c}} > \mathrm{C}_{\mathrm{a}}\) while \(\mathrm{C}_{\mathrm{b}} < \mathrm{C}_{\mathrm{a}}\)
  3. both \(\mathrm{C}_{\mathrm{b}}, \mathrm{C}_{\mathrm{c}} < \mathrm{C}_{\mathrm{a}}\)
  4. \(\mathrm{C}_{\mathrm{a}}=\mathrm{C}_{\mathrm{b}}=\mathrm{C}_{\mathrm{c}}\)

Solution

\(C_a=\frac{\epsilon_o A}{d}\) and \(C_b=\frac{2 \epsilon_o A K}{d(K+1)}\) and \(C_c=\frac{\epsilon_o \frac{A}{2}}{d}+\frac{\epsilon_o \frac{A}{2} K}{d}=\frac{\epsilon_o A}{2 d} 1+K\) or \(C_b=\frac{2 \epsilon_o A}{d} / 1+K > C_a\) or \(C_c=\frac{\epsilon_o A 1+K}{d} \frac{1}{2} > C_a\) \(\therefore C_b\) and \(C_c > C_a\)

Asked in: JEE Mains - Capacitance - Test 1

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