A parallel plate capacitor is to be designed which is to be connected across 1kv potential.The dielectric meteral which is to be Filled between the plates has dielectric constant $K=6\pi$ and dielectric strenght ${10}^{7}V/m.$For safly the eletric field is never to exceed 10 % of the dielectirc strenght.With such specification, of we want a capacitor of 50pF, What minumim area (in $m{m}^2$) of plates is required for safe working ?
$(\text{use}{\epsilon }_0=\frac{1}{36\pi }\times {10}^{-9}\text{in}\text{MKS})$

A parallel plate capacitor is filled with dielectrics. What is capacitance?

A parallel plate capacitor is connected across a source of covstant potential difference. If a dielectric slab is introduced between the two plates, then

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A parallel plate capacitor has two layers of dielectrics as shown in figure. This capacitor is connected across a battery, then the ratio of potential difference across the dielectric layers is

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An uncharged parallel plate capacitor is connected to a battery. The electric field between the plates is 10V/m. Now a dielectric of dielectric constant 2 is inserted between the plates filling the entries space. The electric field between the plates now is

A parallel plate capacitor is charged completely and then disconnected from the battery. IF the separation between the plates is reduced by 50% and the space between the plates if filled with a dielectric slab of dielectric constant 10, then the potential difference between the plates

A parallel plate capacitor is to be designed with a voltage rating 1kV using a material of dielectric constant 10 and dielectric strength ${10}^6\frac{V}{m}$ .What minimum area of the plates is required to have a capacitance of $88.5pF$ ?

A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.

A parallel plate capacitor with air between the plates has a capacitance C. If the distance between the plates is doubled and the space between the plates is filled with a dielectric of dielectric constant 6, then the capacitance will become.