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Karnataka Board PUCPUC Science Class 11

The Separation Between the Plates of a Parallel-plate Capacitor is 0⋅500 Cm and Its Plate Area is 100 Cm2. a 0⋅400 Cm Thick Metal Plate is Inserted into the Gap with Its Faces Parallel to the Plates.

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Question

The separation between the plates of a parallel-plate capacitor is 0⋅500 cm and its plate area is 100 cm2. A 0⋅400 cm thick metal plate is inserted into the gap with its faces parallel to the plates. Show that the capacitance of the assembly is independent of the position of the metal plate within the gap and find its value.

Sum
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Solution

Given:
Area of the plate = 100 cm2 
Separation between the plates = `0.500  "cm" = 5 xx 10^-3  "m"`

Thickness of the metal, `t = 4 xx 10^-3  "m"`

`therefore C = (∈_0A)/(d-t+t/k)`

Here,
k = Dielectric constant of the metal
d = Separation between the plates
t = Thickness of the metal

For the metal, k = ∞

`therefore C = (∈_0A)/(d-t) = ((8.85 xx 10^-12) xx 10^-12)/((5-4) xx 10^-3) = 88  "pF"`

Here, the capacitance is independent of the position of the metal. At any position, the net separation is (d − t).

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Chapter 31: Capacitors - Exercises [Page 169]

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HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 31 Capacitors
Exercises | Q 51 | Page 169

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