(a) A parallel plate capacitor of plate area `2m^(2)` and pltae separation `5xx8.85mm` is cahrged to `200V` in vacuum. Comute the capacitance, charge, charge density and electric field.
(b) The charging is removed and space between the plates is filled with a dielectric `(K=5)`. computer new capacitance, the `p.d`., electric field and induced charge at the surface of dielectric slab.
(c) If now the dielectric sheet is removed and replaced by two sheets, one `2xx8.85mm` thick of dielectric constant `4`, and the other `3xx8.85mm` thick of dielectric constant `2`, compute electric field in each dielectric, the potential difference across the capacitor and its capacitance.

Explanation

(a) `C=(Ain_(0))/(d)=(2xx8.85xx10^(-12))/(5xx8.85xx10^(-3))=4xx10^(-10)F`
`Q=CV=4xx10^(-10)xx200=8xx10^(-8)C`
`sigma=(Q)/(A)=(8xx10^(-8))/(2)=4xx10^(-8)C//m^(2)`
`E=(Q)/(Ain(0))=(4xx10^(-8))/(8.85xx10^(-12))=4500V//m` OR
`E=(V)/(d)=(200)/(5xx8.85xx10^(-3))=4500 V//m`
(b) `C'KC=5xx4xx10^(-10)=2xx10^(-10)=2xx10^(-9)F`
Since cgarge remains same
`Q=CV=C'V'impliesV'=(CV)/(C")=(Q)/(C')=(8xx10^(-8))/(2xx10^(-9))=40V`
`E'=(V')/(d)=(40)/(5xx8.85xx10^(-3))~=900V//m` OR ltbr `E'=(E)/(K)=(4500)/(5)=900V//m`
`Q_(i n)=Q(1-(1)/(K))=(8xx10^(-8))(1-(1)/(5))=6.4xx10^(-8)C`
(c) `W_(1)=(E)/(K_(1))=(4500)/(4)=1125 V//m`
`E_(2)=(E)/(K_(2))=(4500)/(2)=2250V//m`
`C"=(Ain_(0))/(d-(t_(1)+t_(2))+(t_(1))/(K_(1))+(t_(2))/(K_(2)))=(Ain_(0))/((t_(1))/(K_(1))+(t_(2))/(K_(2))`
`=(2xx8.85xx10^(-12))/(((2xx8.85)/(4)+(3xx8.85)/(2))xx10^(-3))=10^(-9)F`
`Q=CV=C"V"`
`V''=(Q)/(C'')=(8xx10^(-8))/(10^(-9))=80V`
OR
`C_(1)=(K_(1)Ain_(0))/(t_(1))=(4xx2xx8.85xx10^(-12))/(2xx8.85xx10^(-3))=4xx10^(-9)F`
`C_(2)=(K_(2)Ain_(0))/(t_(2))=(2xx2xx8.85xx10^(-12))/(3xx8.85xx10^(-3))=(4)/(3)xx10^(-9)F`
`C_(eq)=C=(C_(1)C_(2))/(C_(1)+C_(2))=10^(-9)V`
`V_(1)(C_(2))/(C+_(1)+C_(2))V=20V, V_(2)=V-V_(1)=60V`
`E_(1)=(V_(1))/(t_(1))=(20)/(2xx8.85xx10^(-3))+1125V//m`
`E_(2)=(V_(2))/(t_(2))=(60)/(3xx8.85xx10^(-3)~2250V//m`