$A$ capacitor of capacitance $9 \, nF$ having a dielectric slab of $\varepsilon_{r} = 2.4$,dielectric strength $20 \, MV/m$,and potential difference $V = 20 \, V$. The area of the plates is ....... $\times 10^{-4} \, m^{2}$.

  • A
    $2.1$
  • B
    $4.2$
  • C
    $1.4$
  • D
    $2.4$

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The capacitance of an air-filled parallel plate capacitor is $9 \ pF$. If the space between the plates is filled with two dielectric slabs of thickness $d/3$ with dielectric constant $K_1 = 3$ and thickness $2d/3$ with dielectric constant $K_2 = 6$,the new capacitance will be ...... $pF$.

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$A$ container has a base of $50 \text{ cm} \times 5 \text{ cm}$ and height $50 \text{ cm}$, as shown in the figure. It has two parallel electrically conducting walls each of area $50 \text{ cm} \times 50 \text{ cm}$. The remaining walls of the container are thin and non-conducting. The container is being filled with a liquid of dielectric constant $3$ at a uniform rate of $250 \text{ cm}^3 \text{ s}^{-1}$. What is the value of the capacitance of the container after $10 \text{ s}$ (in $\text{ pF}$)? [Given: Permittivity of free space $\epsilon_0 = 9 \times 10^{-12} \text{ C}^2 \text{ N}^{-1} \text{ m}^{-2}$, the effects of the non-conducting walls on the capacitance are negligible]

$A$ capacitor is connected to a battery of voltage $V$. If a dielectric slab of dielectric constant $k$ is completely inserted between the plates,what will be the final charge on the capacitor? (Assume the initial charge is $q_{0}$)

$A$ parallel plate capacitor with air between the plates has a capacitance of $1.0 \text{ pF}$. If the distance between the plates is doubled and the space between them is filled with a dielectric substance, the capacitance becomes $2.0 \text{ pF}$. Then the value of the dielectric constant of the dielectric substance is . . . . . . .

Which one statement is correct? $A$ parallel plate air condenser is connected to a battery. Its charge,potential,electric field,and energy are ${Q_0}$,${V_0}$,${E_0}$,and ${U_0}$ respectively. $A$ dielectric slab is inserted to fill the complete space between the plates while the battery remains connected. Now,the corresponding values $Q$,$V$,$E$,and $U$ are related to the initial values as:

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