Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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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. - Physics

ConceptCapacitors and Capacitance

#### Question

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.

#### Solution

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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Solution 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. Concept: Capacitors and Capacitance.
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