Question
The two square faces of a rectangular dielectric slab $($dielectric constant $4.0)$ of dimensions $20\ cm \times 20\ cm \times 1.0mm$ are metal$-$coated. Find the capacitance between the coated surfaces.
✓
Answer
$A = 20\ cm \times 20\ cm = 4 \times 10^{-2}m$
$d = 1m = 1 \times 10^{-3}m$
$k = 4$
$t = d$
$\text{C}=\frac{\epsilon_0\text{A}}{\text{d}-\text{t}+\frac{\text{t}}{\text{k}}}=\frac{\epsilon_0\text{A}}{\text{d}-\text{d}+\frac{\text{d}}{\text{k}}}=\frac{\epsilon_0\text{Ak}}{\text{d}}$
$=\frac{8.85\times10^{-12}\times4\times10^{-2}\times4}{1\times10^{-3}}$
$=141.6\times10^{-9}\text{F}=1.42\text{nf}$
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