The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between t

Question

The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric constant 3 and thickness 3 mm is introduced between the plates with its faces parallel to them. The distance between the plates is so adjusted that the capacitance of the capacitor becomes 2/3rd of its original value. What is the new distance between the plates?

TestHub · Accepted Answer

Here, distance between parallel plates d = 4 mm = 0.004 m, K = 3, thickness t = 3 mm = 0.003 m and $d_{1}$ =? $∴ C = \frac{ε_{0} A}{d}$ and $C_{1} = \frac{ε_{0} A}{d_{1} - t (1 - \frac{1}{K})}$ since $C_{1} = \frac{2}{3} C$ (given)
$\begin{array}{l} ∴ \frac{ε_{0} A}{d_{1} - t (1 - \frac{1}{K})} = \frac{2}{3} \frac{ε_{0} A}{d} \\ \frac{1}{d_{1} - t (1 - \frac{1}{K})} = \frac{2}{3 d} \\ \frac{1}{d_{1} - 0.003 (1 - \frac{1}{3})} = \frac{2}{3 \times 0.004} \\ \frac{1}{d_{1} - 0.003 \times \frac{2}{3}} = \frac{1}{0.006} \\ \frac{1}{d_{1} - 0.002} = \frac{1}{0.006} \\ d_{1} - 0.002 = 0.006 \\ d_{1} = 0.006 + 0.002 = 0.008 m = 8 mm . \end{array}$

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