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Electrostatic Potential and Capacitance
Capacitors with Dielectric Slabs
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- Dielectric Slab Capacitance
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2025 1 question
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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:
Using series capacitor formulation with remaining air gap $d/8$, we find $K_2 = 32/15$ and $K_1 = 1.25 \times K_2 = 8/3 \approx 2.66$.
Remaining air thickness is $t_{air} = d - \left(\frac{3d}{8} + \frac{d}{2}\right) = d - \frac{7d}{8} = \frac{d}{8}$. The equivalent capacitance is given by: $\frac{d_{eff}}{\epsilon_0 A} = \frac{t_1}{K_1 \epsilon_0 A} + \frac{t_2}{K_2 \epsilon_0 A} + \frac{t_{air}}{\epsilon_0 A}$. Since $C_{new} = 2 C_0$, $d_{eff} = \frac{d}{2}$. Thus: $\frac{d}{2} = \frac{3d}{8K_1} + \frac{d}{2K_2} + \frac{d}{8} \implies \frac{1}{2} - \frac{1}{8} = \frac{3}{8K_1} + \frac{1}{2K_2} \implies \frac{3}{8} = \frac{3}{8K_1} + \frac{4}{8K_2} \implies 3 = \frac{3}{K_1} + \frac{4}{K_2}$. Given $K_1 = 1.25 K_2 = \frac{5}{4} K_2$: $3 = \frac{3}{(5/4)K_2} + \frac{4}{K_2} = \frac{12}{5K_2} + \frac{20}{5K_2} = \frac{32}{5K_2} \implies K_2 = \frac{32}{15}$. Therefore, $K_1 = \frac{5}{4} \times \frac{32}{15} = \frac{8}{3} \approx 2.667$.