After charging a capacitor the battery is removed. Now by placing a dielectric slab between the plates :-
The potential difference between the plates and the energy stored will decrease but the charge on plates will remain same
the charge on the plates will decrease and the potential difference between the plates will increase
the potential difference between the plates will increase and energy stored will decrease but the charge on the plates will remain same
the potential difference, energy stored and the charge will remain unchanged.
For changing the capacitance of a given parallel plate capacitor, a dielectric material of dielectric constant $K$ is used, which has the same area as the plates of the capacitor. The thickness of the dielectric slab is $\frac{3}{4} d$, where $'d'$ is the separation between the plates of parallel plate capacitor. The new capacitance $(C')$ in terms of original capacitance $\left( C _{0}\right)$ is given by the following relation
Capacitance of a parallel plate capacitor becomes $4/3$ times its original value if a dielectric slab of thickness $t = d/2$ is inserted between the plates ($d$ is the separation between the plates). The dielectric constant of the slab is
The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)
Two thin dielectric slabs of dielectric constants $K_1$ and $K_2$ $(K_1 < K_2)$ are inserted between plates of a parallel plate capacitor, as shown in the figure. The variation of electric field $E$ between the plates with distance $d$ as measured from plate $P$ is correctly shown by
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$