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The Two Square Faces of a Rectangular Dielectric Slab (Dielectric Constant 4⋅0) of Dimensions 20 Cm × 20 Cm × 1⋅0 Mm Are Metal-coated. Find the Capacitance Between the Coated Surfaces.

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प्रश्न

The two square faces of a rectangular dielectric slab (dielectric constant 4⋅0) of dimensions 20 cm × 20 cm × 1⋅0 mm are metal-coated. Find the capacitance between the coated surfaces.

योग
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उत्तर

The area of the plates of the capacitor is given by 

`A = 20  "cm" xx 20  "cm" = 400  "cm"^2`

⇒ `A = 4 xx 10^-2  "m"`

The separation between the parallel plates is given by 

`d = 1  "m" = 1 xx 10^-3  "m"`

Here, the thickness of the dielectric is the same as the separation between the parallel plates.

Thus, the capacitance is given by

`C = (∈_0Ak)/d = ((8.85 xx 10^-12) xx (4 xx 10^-2) xx 4)/10^-3 = 1.42  "nF"`

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अध्याय 31: Capacitors - Exercises [पृष्ठ १६९]

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एचसी वर्मा Concepts of Physics Volume 1 and 2 [English]
अध्याय 31 Capacitors
Exercises | Q 49 | पृष्ठ १६९

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