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# 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. - Physics

Sum

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.

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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).

Concept: Capacitors and Capacitance
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#### APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 9 Capacitors
Q 51 | Page 169
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