Question

Hey!!



A slab of material of dielectric constant K, has the same area as the plates of a parallel plate capacitor, but has thickness d/2 , where d is the separation between the plates , find the expression for the capacitance when the slab is inserted between the plates. ​

Answer 1

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Initially when there is vacuum between the two plates, the capacitance of the capictor is 

C =E°A/d

Where, A is the area of parallel plates 

Suppose that the capacitor is connected to a battery, an electric field 

now if we insert the dielectric slab of thickness t=d/2 the electric field reduce to E

Now, if we insert the dielectric slab of thickness

t=d the electric field reduces to E Now, the gap between plates is divided in two parts, for distance there is electric field E and for the remaining distance

(d-t) the electric field is   E

 If V be the potential difference between the plates of the capacitor, then

$\sf\bold{V={ E }_{ t }+{ E }_{ 0 }(d-t)}$

$\sf\bold{V=\dfrac { Ed }{ 2 } +\dfrac { { E }_{ 0 }d }{ 2 } =\dfrac { d }{ 2 } (E+{ E }_{ 0 })\quad \quad \quad \quad (t=\dfrac { d }{ 2 } )}$

$\sf\bold{V=\dfrac { d }{ 2 } (\dfrac { { E }_{ 0 } }{ K } +{ E }_{ 0 })=\dfrac { d{ E }_{ 0 } }{ 2K } (K+1)\quad \quad \quad \quad (\quad as\quad \dfrac { { E }_{ 0 } }{ E } =K)}$

$\sf\bold{now\:E _{ 0 }=\dfrac { \sigma }{ E_{0}} =\dfrac { q }{ E_{0}}A }$

$\sf\bold{\Rightarrow \:V=\dfrac { d }{ 2K } \dfrac { q }{ E_{ 0 } A } (K+1)}$

$\sf\bold{we\quad know\quad C=\dfrac { q }{ V } =\dfrac { 2K \:E _{ 0 } A }{ (K+1)d }}$