Question

A parallel plate capacitor is made of two circular plates separated by a distance of 5mm and with a dielectric of dielectric constant 2.2 between them. When the electric field in the dielectric is $3\times {{10}^{4}}V{{m}^{-1}}$, the charge density of the positive plate will be close to:
$\text{A}\text{. }3\times {{10}^{4}}C{{m}^{-2}}$
$\text{B}\text{. }6\times {{10}^{4}}C{{m}^{-2}}$
$\text{C}\text{. }6\times {{10}^{-7}}C{{m}^{-2}}$
$\text{C}\text{. 3}\times {{10}^{-7}}C{{m}^{-2}}$

Answer 1

Use the formula $C=k\dfrac{{{\varepsilon }_{0}}A}{d}$ and the find the capacitance of the capacitor. Then find the potential difference across the plates by using V=Ed. Then finally use Q=CV to find the charge density on the positive plate.

Formula used:
$C=k\dfrac{{{\varepsilon }_{0}}A}{d}$
V=Ed
Q=CV

Complete step-by-step answer:
A capacitor is a device that stores charge. We can also say that it stores electrical energy in the form of electric fields. When a potential difference V is produced across the charge (Q) on the positive plates of the capacitor is given as Q = CV.
Here, C is called the capacitance of the capacitor.
Suppose the distance between the parallel plates is d and the area of the plates is A, then the capacitance of the capacitor is equal to $C=\dfrac{{{\varepsilon }_{0}}A}{d}$, where ${{\varepsilon }_{0}}$ is a constant called absolute permittivity.
When we insert a dielectric medium in between the plates, the capacitance of the capacitor increases. And the new value of the capacitance is $C=k\dfrac{{{\varepsilon }_{0}}A}{d}$, where k is the dielectric constant of the dielectric.
It is asked to calculate the charge density on the positive plate. The charge density on the positive plates is equal to the charge on the plate divided by the area of the plate.
i.e. $\dfrac{Q}{A}$
To find $\dfrac{Q}{A}$, first we will be using Q=CV …. (i).
In this case, k=2.2, $d=5mm=5\times {{10}^{-3}}m$ and ${{\varepsilon }_{0}}=8.85\times {{10}^{-12}}\dfrac{{{C}^{2}}}{Nm}$
Hence,
$C=k\dfrac{{{\varepsilon }_{0}}A}{d}=2.2\times \dfrac{8.85\times {{10}^{-12}}.A}{5\times {{10}^{-3}}}=3.9\times {{10}^{-9}}A$
When the distance between the plates is very small, the electric field between the plates is very small. Therefore, the potential difference across the plates becomes V=Ed.
$\Rightarrow V=3\times {{10}^{4}}\times 5\times {{10}^{-3}}=150V$
Substitute the values of C and V in equation (i).
This implies that
$Q=3.9\times {{10}^{-9}}A\times 150$
$\Rightarrow \dfrac{Q}{A}=3.9\times {{10}^{-9}}\times 150=5.8\times {{10}^{-7}}\approx 6\times {{10}^{-7}}C{{m}^{-2}}$
Hence, the correct option is C.

Note: This question can be solved with an alternative method.
The value of electric field between the plates in absence of a dielectric is $E=\dfrac{\sigma }{{{\varepsilon }_{0}}}$ …. (1), where $\sigma$ is the charge density on the positive plate.
When a dielectric is inserted between the plates, the electric field between the plates reduces by some amount. Let the net electric field be E’.
The relation between E and E’ is $E=kE'$. This is how dielectric constant k is defined.
It is given that $E'=3\times {{10}^{4}}V{{m}^{-1}}$ and k=2.2
This gives us that $E=2.2\times 3\times {{10}^{4}}V{{m}^{-1}}=6.6\times {{10}^{4}}V{{m}^{-1}}$
Substitute the value of E from equation (1).
$\Rightarrow \dfrac{\sigma }{{{\varepsilon }_{0}}}=6.6\times {{10}^{4}}V{{m}^{-1}}$
$\Rightarrow \sigma ={{\varepsilon }_{0}}6.6\times {{10}^{4}}=8.85\times {{10}^{-12}}\times 6.6\times {{10}^{4}}\approx 6\times {{10}^{-7}}C{{m}^{-2}}$