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  • I need help with 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 )

  • Option 1)

    \frac{\varepsilon _0A}{\frac{d_1}{K_1}+\frac{d_2}{K_2}+\frac{d_3}{K_3}}

  • Option 2)

    \frac{\varepsilon _0A}{\frac{d_1+d_2+d_3}{K_1+K_2+K_3}}

  • Option 3)

    \frac{\varepsilon _0A({K_1K_2K_3})}

  • Option 4)

    \varepsilon _0\left ( {\frac{AK_1}{d_1}}+\frac{AK_2}{d_2}+\frac{AK_3}{d_3} \right )

 

Answers (1)

best_answer

As we have learned

If dielectric slab Partially filled -

{C}'=\frac{\epsilon _{0}A}{d-t+\frac{t}{k}}

- wherein

 

 

\Rightarrow V'= E\left ( d-t +t/K\right )= \sigma / \varepsilon _0\left ( d-t +t/ K\right )= \frac{Q}{A\varepsilon _0}\left ( d-t + t/K\right )

C'= \frac{\varepsilon _0A}{\frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}+\frac{t_4}{K_4}}  given t= d

 


Option 1)

\frac{\varepsilon _0A}{\frac{d_1}{K_1}+\frac{d_2}{K_2}+\frac{d_3}{K_3}}

Option 2)

\frac{\varepsilon _0A}{\frac{d_1+d_2+d_3}{K_1+K_2+K_3}}

Option 3)

\frac{\varepsilon _0A({K_1K_2K_3})}

Option 4)

\varepsilon _0\left ( {\frac{AK_1}{d_1}}+\frac{AK_2}{d_2}+\frac{AK_3}{d_3} \right )

Posted by

Avinash

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