Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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A Parallel-plate Capacitor Has Plate Area 100 Cm2 and Plate Separation 1⋅0 Cm. a Glass Plate (Dielectric Constant 6⋅0) of Thickness 6⋅0 Mm and an Ebonite Plate (Dielectric Constant 4⋅0) - Physics

ConceptCapacitors and Capacitance

Question

A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.

Solution

The given system of the capacitor will behave as two capacitors connected in series.

Let the capacitances be C1 and C2.
Now,

C_1 = (∈_0Ak_1)/d_1 and C_2 = (∈_0Ak_2)/d_2

Thus, the net capacitance is given by

C = (C_1C_2)/(C_1+C_2)

= ((∈_0Ak_1)/d_1 xx (∈_0Ak_2)/d_2)/((∈_0Ak_1)/d_1 +(∈_0Ak_2)/d_2)

= (∈_0A(k_1+k_2))/(k_1d_2+k_2d_1)

= ((8.85 xx 10^-12) xx (10^-2) xx 24) / ((6 xx 4 xx 10^-3 + 4 xx 6 xx 10^-3)) = 4.425 xx 10^-11 C

= 44.25 pF

Is there an error in this question or solution?

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Solution A Parallel-plate Capacitor Has Plate Area 100 Cm2 and Plate Separation 1⋅0 Cm. a Glass Plate (Dielectric Constant 6⋅0) of Thickness 6⋅0 Mm and an Ebonite Plate (Dielectric Constant 4⋅0) Concept: Capacitors and Capacitance.
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