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Chapter 11: Problem 1616

Capacitance of a parallel plate capacitor becomes \((4 / 3)\) times its original value if a dielectric slab of thickness \(t=d / 2\) is inserted between the plates (d is the separation between the plates). The dielectric constant of the slab is (A) 8 (B) 4 (C) 6 (D) 2

Short Answer

Expert verified
The dielectric constant of the slab is found by analyzing the given information of the capacitance of a parallel plate capacitor and the dielectric slab's thickness. After solving the equations, we find that the dielectric constant of the slab is 8. Therefore, the correct answer is (A) 8.

Step by step solution

01

Understand Capacitance and Dielectric Constant

Capacitance (C) of a parallel plate capacitor depends on the surface area (A) of the plates, the distance (d) between them, and the dielectric constant (k) of the material between the plates. In vacuum or air, k = 1, and the capacitance formula is: C = ε_0 * (A / d), where ε_0 is the vacuum permittivity. When a dielectric slab is introduced between the plates, the capacitance becomes: C' = k * ε_0 * (A / d).
02

Introduce a dielectric slab with thickness t

If we insert a dielectric slab of thickness t = d/2, the separation between the plates can be considered as being made up of two regions: one with the dielectric slab of thickness t and the other being empty (air, dielectric constant k = 1) of thickness (d - t). In this case, the effective capacitance C' is given by the formula for capacitors connected in series: (1 / C') = (1 / C₁) + (1 / C₂), where C₁ is the capacitance of the dielectric-filled region, and C₂ is the capacitance of the air-filled region. From the given information, we can write C' = (4/3)C.
03

Calculate C₁ and C₂

C₁ = K * ε_0 * (A / t), as it is filled with a dielectric slab, and C₂ = ε_0 * (A / (d-t)), as it is filled with air (k=1). Now substitute the values t = d/2, and (d-t) = d/2 in the equations of C₁ and C₂: C₁ = K * ε_0 * (A / (d/2)), C₂ = ε_0 * (A / (d/2)).
04

Find the relation between C' and dielectric constant

Now, using (1 / C') = (1 / C₁) + (1 / C₂), we can write: (1 / ((4/3)C)) = (1 / (K * ε_0 * (A / (d/2)))) + (1 / (ε_0 * (A / (d/2)))). Now, use C = ε_0 * (A / d): 1 / (4/3)C = 1 / (K * 2/1C) + 1 / (2/1C), Let's simplify the equation to find K: 3 = (8 + 4K) / K, 3K = 8 + 4K, K = 8.
05

Choose the correct option

Now, from the options given, we found that the dielectric constant of the slab is: (A) 8. Therefore, the correct answer is (A) 8.

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Most popular questions from this chapter

Two spherical conductors of radii \(\mathrm{r}_{1}\) and \(\mathrm{r}_{2}\) are at potentials \(\mathrm{v}_{1}\) and \(\mathrm{v}_{2}\) respectively, then what will be the common potential when the conductors are brought in constant? (A) $\left[\left(\mathrm{r}_{1} \mathrm{v}_{1}+\mathrm{r}_{2} \mathrm{v}_{2}\right) /\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\right]$ (B) $\left[\left(\mathrm{r}_{1} \mathrm{v}_{1}+\mathrm{r}_{2} \mathrm{v}_{2}\right) /\left(\mathrm{r}_{1}-\mathrm{r}_{2}\right)\right]$ (C) $\left[\left(\mathrm{r}_{1} \mathrm{v}_{1}-\mathrm{r}_{2} \mathrm{v}_{2}\right) /\left(\mathrm{r}_{1}+\mathrm{r}_{2}\right)\right]$ (D) None of these

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