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Topic: Capacitors, series and parallel combination, parallel plate capacitor with/without dielectric· Electrostatics

The plates of a parallel plate capacitor are separated by d. Two slabs of different dielectric constant K_1 and K_2 with thickness 3/8d and d/2, respectively are inserted in the capacitor. Due to this, the capacitance becomes two times larger than wh…

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

1 answer

Best answer

Answer: 1. 2.66

Key idea: for slabs inserted in a capacitor, \(C=\dfrac{\varepsilon_0A}{d-t_1-t_2+t_1/K_1+t_2/K_2}\).

Here \(t_1=\tfrac38d\), \(t_2=\tfrac12d\), and \(C=2C_0\), so the denominator must equal \(d/2\):

$$d-\frac38d-\frac12d+\frac{3d}{8K_1}+\frac{d}{2K_2}=\frac d2\Rightarrow\frac{3}{8K_1}+\frac{1}{2K_2}=\frac38$$

With \(K_2=K_1/1.25=0.8K_1\): \(\dfrac{1}{2K_2}=\dfrac{0.625}{K_1}\). So

$$\frac{0.375+0.625}{K_1}=\frac38\Rightarrow K_1=\frac83\approx2.67$$
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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