Question: 20. The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K_{1} and K_{2} with thickness \frac{3}{8} d and \frac{d}{2}, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than when there is nothing between the plates. If K_{1}=1.25 K_{2}, the value of K_{1} is:
(1) 2.66
(2) 2.33
(3) 1.60
(4) 1.33
Answer: Option (1)
Explanation:
When dielectric slabs are placed in series between capacitor plates, the effective capacitance is obtained by adding the dielectric thickness ratios:
\frac{d_{\text{eff}}}{d}=\frac{3}{8K_{1}}+\frac{1}{2K_{2}}+\frac{1}{8}The capacitance becomes twice the original value, so
\frac{d}{d_{\text{eff}}}=2 \quad \Rightarrow \quad d_{\text{eff}}=\frac{d}{2}Thus,
\frac{1}{2}=\frac{3}{8K_{1}}+\frac{1}{2K_{2}}+\frac{1}{8}Given that K_{1}=1.25K_{2}, let K_{2}=k.
Then K_{1}=1.25k.
Substitute:
\frac{1}{2}=\frac{3}{8(1.25k)}+\frac{1}{2k}+\frac{1}{8}Simplify terms:
\frac{3}{8(1.25k)}=\frac{3}{10k} \frac{1}{2k}=\frac{5}{10k}Thus,
\frac{1}{2}=\frac{3}{10k}+\frac{5}{10k}+\frac{1}{8} \frac{1}{2}=\frac{8}{10k}+\frac{1}{8}Subtract \frac{1}{8} from both sides:
\frac{1}{2}-\frac{1}{8}=\frac{3}{8}=\frac{8}{10k}Thus,
\frac{8}{10k}=\frac{3}{8} k=\frac{8 \cdot 8}{3 \cdot 10}=\frac{64}{30}=\frac{32}{15}=2.13Therefore,
K_{1}=1.25k=1.25 \times 2.13=2.66Hence the value of K_{1} is 2.66.