The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)
The expression for the capacity of the capacitor formed by compound dielectric placed between the plates of a parallel plate capacitor as shown in figure, will be (area of plate $ = A$)
- A
$\frac{{{\varepsilon _0}A}}{{\left( {\frac{{{d_1}}}{{{K_1}}} + \frac{{{d_2}}}{{{K_2}}} + \frac{{{d_3}}}{{{K_3}}}} \right)}}$
- B
$\frac{{{\varepsilon _0}A}}{{\left( {\frac{{{d_1} + {d_2} + {d_3}}}{{{K_1} + {K_2} + {K_3}}}} \right)}}$
- C
$\frac{{{\varepsilon _0}A({K_1}{K_2}{K_3})}}{{{d_1}{d_2}{d_3}}}$
- D
${\varepsilon _0}\left( {\frac{{A{K_1}}}{{{d_1}}} + \frac{{A{K_2}}}{{{d_2}}} + \frac{{A{K_3}}}{{{d_3}}}} \right)$
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A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :
A source of potential difference $V$ is connected to the combination of two identical capacitors as shown in the figure. When key ' $K$ ' is closed, the total energy stored across the combination is $E _{1}$. Now key ' $K$ ' is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now $E _{2}$. The ratio $E _{1} / E _{2}$ will be :
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