The plates of a parallel plate capacitor of capacity $C_1$ are moved closer together until they are half their original separation. The new capacitance $C_2$ is

  • A
    $C_2 = \frac{C_1}{2}$
  • B
    $C_2 = C_1$
  • C
    $C_2 = 2C_1$
  • D
    $C_2 = 4C_1$

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If the plates of a parallel plate capacitor connected to a battery are moved close to each other,then
$A$. the charge stored in it,increases.
$B$. the energy stored in it,decreases.
$C$. its capacitance increases.
$D$. the ratio of charge to its potential remains the same.
$E$. the product of charge and voltage increases.
Choose the most appropriate answer from the options given below:

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$A$ parallel plate capacitor is to be designed,using a dielectric of dielectric constant $5$,so as to have a dielectric strength of $10^9 \; Vm^{-1}$. If the voltage rating of the capacitor is $12 \; kV$,the minimum area of each plate required to have a capacitance of $80 \; pF$ is

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