The plates of parallel plate capacitor are charged upto $100\;V$. A $2\,mm$ thick plate is inserted between the plates. Then to maintain the same potential difference, the distance between the plates is increased by $1.6\;mm$. The dielectric constant of the plate is
$5$
$1.25$
$4$
$2.5$
In one design of capacitor thin sheets ot metal of area $80\ mm \times 80\ mm$ sandwich between them a piece of paper whose thickness is $40\ μm$. The relative permittivity of the paper is $4.0$ and its dielectric strength is $20\ MVm^{-1}$. Calculate the maximum charge that can be put on the capacitor
[permittivity of free space $ = 9 \times 10^{-12}\ Fm^{-1}$]
If a slab of insulating material $4 \times {10^{ - 3}}\,m$ thick is introduced between the plates of a parallel plate capacitor, the separation between plates has to be increased by $3.5 \times {10^{ - 3}}\,m$ to restore the capacity to original value. The dielectric constant of the material will be
The radii of the inner and outer spheres of a condenser are $9\,cm$ and $10\,cm$ respectively. If the dielectric constant of the medium between the two spheres is $6$ and charge on the inner sphere is $18 \times {10^{ - 9}}\;coulomb$, then the potential of inner sphere will be, if the outer sphere is earthed........$volts$
A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage $(U )$ as $\varepsilon = \alpha U$ where $\alpha = 2{V^{ - 1}}$. A similar capacitor with no dielectric is charged to ${U_0} = 78\,V$. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
A parallel plate air-core capacitor is connected across a source of constant potential difference. When a dielectric plate is introduced between the two plates then :