In a parallel plate capacitor the distance between the plates is 10 cm. Two dielectric constants ${K_1} = 2$ and ${K_2} = 4$ respectively, are inserted between the plates. A potential of 100V is applied across the capacitor as shown in the figure. The value of the net bound surface charge density at the interface of the two dielectrics is:
A. $ - \dfrac{{2000}}{3}{\varepsilon _ \circ }$
B. $ - \dfrac{{1000}}{3}{\varepsilon _ \circ }$
C. $ - 250{\varepsilon _ \circ }$
D. $\dfrac{{2000}}{3}{\varepsilon _ \circ }$

Hint: Surface charge density is defined as the measure of amount of electric charge is accumulated over a surface. In other words, it is a quantity of charge measured per unit surface area. If ‘q’ is the charge and ‘A’ is the area of the surface, then the surface charge density is written as, $\sigma = qA$. The SI unit of the surface charge density is expressed in $C/{m^2}$
Formula Used: $V = Ed$

Complete answer:
A parallel plate capacitor is set up by an arrangement of electrodes and a dielectric or an insulating material. It can only store a finite amount of energy before any dielectric breakdown.
When two parallel plates are connected across a battery, the plates are charged and an electric field is established between them, this setup is known as the parallel plate capacitor. In the given parallel plate capacitor with the given distance between the plates the charge density can be given as, $\sigma = \dfrac{Q}{A}$.
V = ${E_1}d + {E_2}d = \dfrac{\sigma }{{{\varepsilon _1}}}d + \dfrac{\sigma }{{{\varepsilon _2}}}d = \dfrac{\sigma }{{2{\varepsilon _ \circ }}}d + \dfrac{\sigma }{{4{\varepsilon _ \circ }}}d = \dfrac{{3\sigma }}{{4{\varepsilon _ \circ }}}d$
It is given that the V = 100 Volts, d = $5 \times {10^{ - 2}}cm$
$ \Rightarrow \sigma = \dfrac{{4{\varepsilon _ \circ }}}{{3d}}V = \dfrac{{4{\varepsilon _ \circ }}}{{3 \times 5 \times {{10}^{ - 2}}}} \times 100 = \dfrac{{4 \times {{10}^4}}}{{15}}{\varepsilon _ \circ }$
$\overline {{P_1}} = {\varepsilon _ \circ }x\overrightarrow {{E_1}} = {\varepsilon _ \circ }\left( {{K_2} - 1} \right)\overrightarrow {{E_1}} $
$ \Rightarrow {\sigma _1} = {\varepsilon _ \circ } \times \dfrac{\sigma }{{2{\varepsilon _ \circ }}} = \dfrac{\sigma }{2}$
${\overline P _2} = {\varepsilon _ \circ }x\overrightarrow {{E_2}} = {\varepsilon _ \circ }\left( {{K_2} - 1} \right)\overrightarrow {{E_2}} $
$ \Rightarrow {\sigma _1} = 3{\varepsilon _ \circ } \times \dfrac{\sigma }{{4{\varepsilon _ \circ }}} = \dfrac{{3\sigma }}{4}$
Now, $\sigma = {\sigma _1} - {\sigma _2} = \dfrac{\sigma }{2} - \dfrac{{3\sigma }}{4} = - \dfrac{\sigma }{4}$
$ \Rightarrow \sigma = - \dfrac{1}{4} \times \dfrac{{4 \times {{10}^4}}}{{15}}{\varepsilon _ \circ } = - \dfrac{{2000}}{3}{\varepsilon _ \circ }$
Thus, the value of the net bound surface charge density at the interface of the two dielectrics is $ - \dfrac{{2000}}{3}{\varepsilon _ \circ }$.

Hence, option (A) is the correct answer.

Note:
The electric potential is defined as the amount of work done in order to move a unit charge from a reference point to a specific point against an electric field. When two plates consist of vacuum between each other, then the potential energy across the capacitor is given by, $V = Ed$; where ‘E’ is denoted as a uniform electric field and ‘d’ is expressed as the distance between the two-point charges.