(B) Given: Thickness of mica $d_1 = 1 \times 10^{-3} \,m$, thickness of fiber $d_2 = 0.5 \times 10^{-3} \,m$. Dielectric constants $K_1 = 8$ (mica) an

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(B) Given: Thickness of mica $d_1 = 1 \times 10^{-3} \,m$, thickness of fiber $d_2 = 0.5 \times 10^{-3} \,m$. Dielectric constants $K_1 = 8$ (mica) and $K_2 = 2.5$ (fiber). Breakdown electric field of fiber $E_2 = 6.4 \times 10^6 \,V/m$.Since the sheets are in series, the electric displacement field $D = \epsilon_0 K E$ is constant across the plates, so $K_1 E_1 = K_2 E_2$.Thus, $E_1 = E_2 (K_2 / K_1) = (6.4 \times 10^6) \times (2.5 / 8) = 2.0 \times 10^6 \,V/m$.The maximum voltage $V_{max}$ is given by $V_{max} = E_1 d_1 + E_2 d_2$.$V_{max} = (2.0 \times 10^6 \times 1 \times 10^{-3}) + (6.4 \times 10^6 \times 0.5 \times 10^{-3})$.$V_{max} = 2000 + 3200 = 5200 \,V$.

Class 12 Physics Electrostatic Potential and Capacitance Effect of Dielectric Inside Capacitor

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The space between the plates of a parallel plate capacitor is filled with a mica sheet of thickness $1 \times 10^{-3} \,m$ and a fiber sheet of thickness $0.5 \times 10^{-3} \,m$. The dielectric constants of mica and fiber are $8$ and $2.5$ respectively. If the fiber breaks down at an electric field of $6.4 \times 10^6 \,V/m$, then the maximum voltage that can be applied to the capacitor is: (in $\,V$)

AP EAMCET 2023 All AP EAMCET

A
$3400$
B
$5200$
C
$2700$
D
$4800$

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Class 12

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Electrostatic Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ parallel plate capacitor has plates of length $l$,width $w$,and separation $d$. It is connected to a battery of emf $V$. $A$ dielectric slab of the same thickness $d$ and dielectric constant $k = 4$ is being inserted between the plates. At what length $x$ of the slab inside the plates will the energy stored in the capacitor be two times the initial energy stored?

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$A$ $60 \ \mu F$ parallel plate capacitor whose plates are separated by $6 \ mm$ is charged to $250 \ V$,and then the charging source is removed. When a slab of dielectric constant $5$ and thickness $3 \ mm$ is placed between the plates,find the change in the potential difference across the capacitor (in $V$)?

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The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

Difficult

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In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

Medium

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$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

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The plates of a parallel plate capacitor are charged up to $100\,V$. $A$ $2\,mm$ thick dielectric plate is inserted between the plates. To maintain the same potential difference,the distance between the capacitor plates is increased by $1.6\,mm$. The dielectric constant of the plate is:

In a parallel plate capacitor of capacitance $C$,a metal sheet is inserted between the plates,parallel to them. If the thickness of the sheet is half of the separation between the plates,the new capacitance will be:

$A$ capacitor of capacitance $C = 10 \ \mu F$ is connected to a $12 \ V$ battery. When a dielectric slab of dielectric constant $K = 5$ is inserted between its plates,how much additional charge (in $\mu C$) will flow from the battery to the capacitor?

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