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Capacitance of a Random Conductor

We know capac...

Question

We know capacitance of capacitor is constant, so why capacitance increase by k factor when dielectric slab of dielectric constant k is inserted?

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The Figure shows capacitors

A

B

C

each. A dielectric slab of dielectric constant

k = 2

and same dimensions as the capacitors is inserted between the plates of the capacitor

B

. In the process of inserting the dielectric slab:

Q. In a capacitor of capacitance

20 μ F

the distance between the plates is

2 m m .

If a dielectric slab of width

1 m m

and dielectric constant

2

is inserted between the plates, then the new capacitance will be:

Q. In given circuit,

C_{1} = C_{2} = C_{3} = C

initially. Now, a dielectric slab of dielectric constant

K = \frac{3}{2}

C_{2}

.
The equivalent capacitance become

Q. In a capacitor of capacitance

20 μ F

, the distance between the plates is

2 mm

. If a dielectric slab of width

1 mm

and dielectric constant

2

is inserted between the plates, then the new capacitance will be :

Q. The capacitance of a parallel plate capacitor is

C_{a}

(fig). A dielectric of dielectric constant K is inserted as shown in fig. If

C_{b}

C_{c}

denote that capacitances in fig, then

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Capacitance of a Random Conductor

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