(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 Electric 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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Chapter

Electric Potential and Capacitance

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Effect of Dielectric Inside Capacitor

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$A$ parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. $A$ dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor,is now inserted in it. Which of the following is incorrect $?$

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The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

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The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

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When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

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$A$ slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$,where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be (Given $C_{0} =$ capacitance of capacitor with air as medium between plates):

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$A$ parallel plate air capacitor of capacitance $C$ is connected to a cell of emf $V$ and then disconnected from it. $A$ dielectric slab of dielectric constant $K,$ which can just fill the air gap of the capacitor,is now inserted in it. Which of the following is incorrect $?$

The capacitance of an air capacitor is $15\,\mu F$ and the separation between the parallel plates is $6\,mm$. $A$ copper plate of $3\,mm$ thickness is introduced symmetrically between the plates. The capacitance now becomes.........$\mu F$.

The distance between the plates of a parallel plate capacitor is $8\,mm$ and the potential difference $(P.D.)$ is $120\,V$. If a $6\,mm$ thick slab of dielectric constant $K = 6$ is introduced between its plates,then:

When the air between the plates of a charged parallel plate capacitor is replaced by a dielectric medium,the intensity of the electric field:

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