Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • Stuck here, help me understand: 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 a number of dielectric slab inserted between the Plate -

{C}'=\frac{\epsilon _{0}A}{d-\left ( t_{1}+t_{2} +\cdots \right )+\left ( \frac{R_{1}}{k_{1}}+\frac{R_{2}}{k_{2}}+\cdots \right )}

- wherein

 

 

C'= \frac{\varepsilon _0A}{\left ( \frac{t_1}{K_1}+\frac{t_2}{K_2}+\frac{t_3}{K_3}+\frac{t_4}{K_4} \right )}

 


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

Plabita

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026: JCECEB invites fresh applications for Jharkhand state merit list correction

Latest Question