JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
The gap between the plates of a parallel plate capacitor of area \(A\) and distance between plates \(d\), is filled with a dielectric whose permittivity varies linearly from \({ \varepsilon _1}\) at one plate to \({ \varepsilon _2}\) at the other. The capacitance of capacitor is
- A \({ \varepsilon _0}\left( {{ \varepsilon _1} + { \varepsilon _2}} \right)A/d\)
- B \({ \varepsilon _0}\left( {{ \varepsilon _2} + { \varepsilon _1}} \right)A/2d\)
- C \({ \varepsilon _0}\,A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]\)
- D \({ \varepsilon _0}\left( {{ \varepsilon _2} - { \varepsilon _1}} \right)A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]\)
Answer & Solution
Correct Answer
(D) \({ \varepsilon _0}\left( {{ \varepsilon _2} - { \varepsilon _1}} \right)A/\left[ {d\,\ln \left( {{ \varepsilon _2}/{ \varepsilon _1}} \right)} \right]\)
Step-by-step Solution
Detailed explanation
As the permittivity of dielectric varies linearly from \(\varepsilon_{1}\) at one plate to \(\varepsilon_{2}\) at the other, it is governed by equation, \(k=\left(\frac{\varepsilon_{2}-\varepsilon_{1}}{d}\right) x+\varepsilon_{1}\) Consider a small element of thickness \(d x\)…
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