If q_{f} is the free charge on the capacitor | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) When a dielectric slab is placed between the plates of a capacitor,the net electric field $E$ inside the dielectric is reduced by a factor of $k$

Vedclass · Accepted Answer

$q_{b} = q_{f} \left(1 - \frac{1}{k}\right)$

Vedclass · Answer

(A) When a dielectric slab is placed between the plates of a capacitor,the net electric field $E$ inside the dielectric is reduced by a factor of $k$ compared to the initial electric field $E_{0}$ in vacuum.The net electric field is given by $E = E_{0} - E_{b}$,where $E_{b}$ is the electric field due to the induced (bound) charges.Since $E = \frac{E_{0}}{k}$,we have $\frac{E_{0}}{k} = E_{0} - E_{b}$.This implies $E_{b} = E_{0} - \frac{E_{0}}{k} = E_{0} \left(1 - \frac{1}{k}\right)$.Since the electric field is proportional to the surface charge density $(E = \frac{\sigma}{\epsilon_{0}})$,the bound charge $q_{b}$ is related to the free charge $q_{f}$ by the same factor:$q_{b} = q_{f} \left(1 - \frac{1}{k}\right)$.

If $q_{f}$ is the free charge on the capacitor plates and $q_{b}$ is the bound charge on the dielectric slab of dielectric constant $k$ placed between the capacitor plates,then the bound charge $q_{b}$ can be expressed as:

Similar Questions

$A$ parallel plate capacitor has plates of area $0.4 \pi \,m^2$ and spacing of $0.5 \,mm$. If a slab of thickness $0.5 \,mm$ and dielectric constant $4.5$ is introduced in between the plates of the capacitor, then the capacitance of the capacitor is

Two identical parallel plate air capacitors are connected in series to a battery of emf $V$. If one of the capacitors is completely filled with a dielectric material of constant $K$,then the potential difference across the other capacitor will become:

Vedclass Test Series

Exam Paper Generator

Online Exam Module