If the dielectric constant and dielectric strength be denoted by $k$ and $x$ respectively, then a material suitable for use as a dielectric in a capacitor must have
High $k$ and high $x$
High $k$ and low $x$
Low $k$ and low $x$
Low $k$ and high $x$
A combination of parallel plate capacitors is maintained at a certain potential difference When a $3\, mm$ thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by $2.4\, mm$. Find the dielectric constant of the slab.
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}$]
A parallel plate capacitor is made of two circular plates separated by a distance $5\ mm$ and with a dielectric of dielectric constant $2.2$ between them. When the electric field in the dielectric is $3 \times 10^4$ $ Vm^{-1}$ the charge density of the positive plate will be close to
A parallel plate capacitor has plates of area $A$ separated by distance $d$ between them. It is filled with a dielectric which has a dielectric constant that varies as $\mathrm{k}(\mathrm{x})=\mathrm{K}(1+\alpha \mathrm{x})$ where $\mathrm{x}$ is the distance measured from one of the plates. If $(\alpha \text {d)}<<1,$ the total capacitance of the system is best given by the expression
Two parallel plates have equal and opposite charge. When the space between them is evacuated, the electric field between the plates is $2 \times {10^5}\,V/m$. When the space is filled with dielectric, the electric field becomes $1 \times {10^5}\,V/m$. The dielectric constant of the dielectric material