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

  • A

    $6$

  • B

    $8$

  • C

    $10$

  • D

    $12$

Similar Questions

Match the pairs

Capacitor Capacitance
$(A)$ Cylindrical capacitor $(i)$ ${4\pi { \in _0}R}$
$(B)$ Spherical capacitor $(ii)$ $\frac{{KA{ \in _0}}}{d}$
$(C)$ Parallel plate capacitor having dielectric between its plates $(iii)$ $\frac{{2\pi{ \in _0}\ell }}{{ln\left( {{r_2}/{r_1}} \right)}}$
$(D)$ Isolated spherical conductor $(iv)$ $\frac{{4\pi { \in _0}{r_1}{r_2}}}{{{r_2} - {r_1}}}$

Explain polarization of nonpolar molecule in uniform electric field and define the linear isotropic dielectrics.

Dielectric constant for metal is

A parallel plate capacitor with plate area $A$ and plate separation $d =2 \,m$ has a capacitance of $4 \,\mu F$. The new capacitance of the system if half of the space between them is filled with a dielectric material of dielectric constant $K =3$ (as shown in figure) will be .........$ \mu \,F$

  • [JEE MAIN 2022]

A parallel plate capacitor with air between the plates has a capacitance of $9\,pF$. The separation between its plates is $'d'$. The space between the plates is now filled with two dielectrics. One of the dielectrics has dielectric constant $K_1=3$ and thickness $\frac {d}{3}$ while the other one has dielectric constant $K_2 = 6$ and thickness $\frac {2d}{3}$ . Capacitance of the capacitor is now........$pF$