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 distance between plates of a parallel plate capacitor is $5d$. Let the positively charged plate is at $ x=0$ and negatively charged plate is at $x=5d$. Two slabs one of conductor and other of a dielectric of equal thickness $d$ are inserted between the plates as shown in figure. Potential versus distance graph will look like :

When a dielectric material is introduced between the plates of a charges condenser, then electric field between the plates

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……..$volts$

The distance between the plates of a parallel plate condenser is $8\,mm$ and $P.D.$ $120\;volts$. If a $6\,mm$ thick slab of dielectric constant $6$ is introduced between its plates, then