Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

 

 Separation between the plates of a parallel plate capacitor is d and the area of each plate is A . When a slab of material of dielectric constant k and thickness t(t < d) is introduced between the plates,  its capacitance become

  • Option 1)

    \frac{\varepsilon _0A}{d+t\left ( 1-\frac{1}{k} \right )}

  • Option 2)

    \frac{\varepsilon _0A}{d+t\left ( 1+\frac{1}{k} \right )}

  • Option 3)

    \frac{\varepsilon _0A}{d-t\left ( 1-\frac{1}{k} \right )}

  • Option 4)

    \frac{\varepsilon _0A}{d-t\left ( 1+\frac{1}{k} \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

 

 

Potential difference between the plates V = Vair + Vmedium

 

 

= \frac{\sigma }{\varepsilon _0}\times (d-t)+\frac{\sigma }{K\varepsilon _0}\times t

 

ÞV= \frac{\sigma }{\varepsilon _0}(d-t+\frac{t}{K})

= \frac{Q}{A\varepsilon _0}(d-t+t/k)

Hence capacitance C=\frac{Q}{V}=\frac{Q}{\frac{Q}{A\varepsilon _0}(d-t+\frac{t}{K})}

= \frac{\varepsilon _0A}{(d-t+t/k)}=\frac{\varepsilon _0A}{(d-t)(1-1/k)}

 


Option 1)

\frac{\varepsilon _0A}{d+t\left ( 1-\frac{1}{k} \right )}

Option 2)

\frac{\varepsilon _0A}{d+t\left ( 1+\frac{1}{k} \right )}

Option 3)

\frac{\varepsilon _0A}{d-t\left ( 1-\frac{1}{k} \right )}

Option 4)

\frac{\varepsilon _0A}{d-t\left ( 1+\frac{1}{k} \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 Result 2024 Live: JEE Mains session 2 results at jeemain.nta.ac.in soon; rank predictor
anam-khan

JEE Main 2024 session 2 final answer key out at jeemain.nta.ac.in; steps to check probable scores
anam-khan

JEE Main Result 2024 Live: Download session 2 final answer key at jeemain.nta.ac.in; rank predictor, cutoff

Latest Question