Question

Electric field inside the capacitor is $E$ and the dielectric constant of the material is $k$. Find the charge density $\sigma $ on the plates. Given $E=6\times {{10}^{5}}V{{m}^{-1}},k=6$.

Answer 1

Hint: Use the formula for the electric field between the plates of a parallel plate capacitor in presence of a dielectric material. Substitute the given values in the formula and calculate the value of the charge density on the plates.

Formula used:
$E=\dfrac{\sigma }{k{{\varepsilon }_{0}}}$
where E is the electric field between the parallel plates of the capacitor, $\sigma $ is the charge density on the plates, k is the dielectric constant of the material and ${{\varepsilon }_{0}}$ is the permittivity of free space.

Complete step by step answer:
Suppose a parallel plate capacitor is connected to an energy source (a battery) and is completely charged. Due to the charges on the plates of the capacitor, an electric field is produced inside the capacitor or in between the plates of the parallel plate capacitor.However, when a dielectric material is placed inside a capacitor or in between the plates of a parallel plate capacitor, the electric field in that space reduces.

Charge density on the plates is equal to charge per unit area of the plate. The electric field in a parallel plate capacitor in presence of a dielectric material is given as,
$E=\dfrac{\sigma }{k{{\varepsilon }_{0}}}$ …. (i)
It is given that the electric field between the parallel plates is equal to,
$E=6\times {{10}^{5}}V{{m}^{-1}}$
The dielectric constant of the dielectric material is equal to $k=6$. The value of permittivity of free space is given to be,
${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}{{N}^{-1}}{{C}^{2}}{{m}^{-2}}$.
Substitute the known values in equation (i).
$\Rightarrow 6\times {{10}^{5}}=\dfrac{\sigma }{6\times 8.85\times {{10}^{-12}}}$
$\therefore \sigma =6\times {{10}^{5}}\times 6\times 8.85\times {{10}^{-12}}=318.6\times {{10}^{-7}}C{{m}^{-2}}$.

This means that the charge density on the plates of the capacitor is equal to $318.6\times {{10}^{-7}}C{{m}^{-2}}$.

Note: In some books you may find the formula for the electric field between the parallel plates in presence of a dielectric material given to be,
$E=\dfrac{\sigma }{\varepsilon }$
Here, $\varepsilon $ is the permittivity of the dielectric medium. The permittivity of the dielectric medium is given as $\varepsilon =k{{\varepsilon }_{0}}$.