Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
Two dielectric slabs of constant ${K_1}$ and ${K_2}$ have been filled in between the plates of a capacitor as shown below. What will be the capacitance of the capacitor
- A
$\frac{{2{\varepsilon _0}A}}{2}({K_1} + {K_2})$
- B
$\frac{{2{\varepsilon _0}A}}{2}\left( {\frac{{{K_1} + {K_2}}}{{{K_1} \times {K_2}}}} \right)$
- C
$\frac{{2{\varepsilon _0}A}}{2}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)$
- D
$\frac{{2{\varepsilon _0}A}}{d}\left( {\frac{{{K_1} \times {K_2}}}{{{K_1} + {K_2}}}} \right)$
Similar Questions
Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A$, separated by a distance $d$. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants $K_1$ , $K_2$ and $K_3$ . The first capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$, the ratio of the energy stored in the two, would be ( $E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$)
Two identical parallel plate capacitors, of capacitance $C$ each, have plates of area $A$, separated by a distance $d$. The space between the plates of the two capacitors, is filled with three dielectrics, of equal thickness and dielectric constants $K_1$ , $K_2$ and $K_3$ . The first capacitor is filled as shown in fig. $I$, and the second one is filled as shown in fig. $II$. If these two modified capacitors are charged by the same potential $V$, the ratio of the energy stored in the two, would be ( $E_1$ refers to capacitor $(I)$ and $E_2$ to capacitor $(II)$)
- [JEE MAIN 2019]
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