Question

11)derive the expression for capacity of parallel plate capacitor when dielectric material is partial filled??​

Answer 1

Answer:

You can mould this expression a little for the required expression... this is the basic one.

Answer 2

Consider a parallel plate capacitor with a di-electric slab such that,

➟ $\pm$q = charge on each plate.

➟ A = Area of plate

➟ d = seperation between the plates.

➟ K = di-electric constant

➟ t = thickness of the plate

➟ $\sigma$ = $\dfrac{q}{A}$

Now,

$\implies \: V\: = \: E_{o}(d \: - \: t) \: + \: E_{m}t$

$\implies \: V \: = \: E_{o}(d \: - \: t) \: + \: \dfrac{E_{o}}{K}t$

$\implies \: V \: = \: E_{o}(d \: - \: t \: + \: \frac{t}{K} )$

$\implies \: V \: = \: \dfrac{ \sigma}{ \epsilon_{o}}(d \: - \: t \: + \: \frac{t}{K} )$

$\implies \: V \: = \: \dfrac{ q}{ \epsilon_{o}A}(d \: - \: t \: + \: \frac{t}{K} )$

Now,

$\Rightarrow\:C\:=\:\dfrac{q}{V}$

$\implies \: C \: = \: \dfrac{ q\epsilon_{o}A}{q} \: \times \: \dfrac{1}{(d \: - \: t \: + \: \frac{t}{K} )}$

$\implies \: C \: = \: \dfrac{ \epsilon \: A}{(d \: - \: t \: + \: \frac{t}{K} )}$

$\implies \: C \: = \: \dfrac{ \epsilon_{o}A }{d(1 \: - \: \frac{t}{d} \: + \: \frac{t}{Kd} )}$

$\implies \: C \: = \: \dfrac{ C_{o} }{(1 \: - \: \frac{t}{d} \: + \: \frac{t}{Kd} )}$

∴ $C\:> \:C_{o}$

Thus, capacitance of parallel plate capacitor increases on inserting di-electric slab between two parallel plates.

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✯ Capacitor :

It is a device which is used to store electrical charge and release it at once.

=> q $\alpha$ V

=> q = CV

=> C = q/V

Here.. C is independent of q or V. But depends on shape and nature of the material.