A parallel plate air filled capacitor shown in Fig.(a) has a capacitance of \(2 \mu \mathrm{F}\). When it is…

A parallel plate air filled capacitor shown in Fig.(a) has a capacitance of \(2 \mu \mathrm{F}\). When it is half filled with a dielectric of dielectric constant \(k=3\) as shown in Fig.(b), its capacitance becomes
  1. \(4 \mu \mathrm{F}\)
  2. \(3 \mu \mathrm{F}\)
  3. \(1.5 \mu \mathrm{F}\)
  4. \(0.5 \mu \mathrm{F}\)

Solution

If \(A\) is the area of each plate, the capacitance of the air-filled capacitor shown in Fig. 11.53(a) on page \(11.26\) is
$C_{0}=\frac{\varepsilon_{0} A}{d}$, where $C_{0}=2 \mu F$ (given). The capacitance of air capacitor in Fig. \(11.54\) (b) is
\(C_{1}=\frac{\varepsilon_{0} A / 2}{d}=\frac{\varepsilon_{0} A}{2 d}=\frac{C_{0}}{2}\)
The capacitance of dielectric filled capacitor in Fig. \(11.53(\mathrm{~b})\) is
\(C_{2}=\frac{k \varepsilon_{0} A / 2}{d}=\frac{k \varepsilon_{0} A}{2 d}=\frac{k C_{0}}{2}\)
Since \(C_{1}\) and \(C_{2}\) are in parallel, the capacitance \(C\) of the capacitor shown in Fig. (b) is
$\begin{aligned} C &=C_{1}+C_{2}=\frac{C_{0}}{2}+\frac{k C_{0}}{2} \\ &=\frac{C_{0}}{2}(1+k)=\frac{2 \mu \mathrm{F}}{2}(1+3)=4 \mu \mathrm{F} \end{aligned}$ Hence the correct choice is (a).

Asked in: JEE Mains - Capacitance - Test 2

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