kanwal32
  • kanwal32
Equivalent capacitance dielectric (if possible tell me what am i doing wrong
Physics
jamiebookeater
  • jamiebookeater
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kanwal32
  • kanwal32
|dw:1434454283651:dw|
kanwal32
  • kanwal32
|dw:1434454418404:dw|
kanwal32
  • kanwal32
lower portion of the capacitance are in parralel so Ceq=\[\frac{ 2EoA/2 }{ d/2 }+\frac{ 4EoA/2 }{ d/2 }\]

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kanwal32
  • kanwal32
EoA/d=C
kanwal32
  • kanwal32
2C+4C=Ceq
kanwal32
  • kanwal32
6c=Ceq
kanwal32
  • kanwal32
now the lower and upper one are in series
kanwal32
  • kanwal32
\[\frac{ d/2 }{ 6kEoA }+\frac{ 1 }{ 6C }\]
kanwal32
  • kanwal32
\[\frac{ 3 }{ 12C }\]=1/Ceq
kanwal32
  • kanwal32
Ceq=4C but my book says 3.9
kanwal32
  • kanwal32
3.9C
kanwal32
  • kanwal32
kanwal32
  • kanwal32
Michele_Laino
  • Michele_Laino
using the law: \[\Phi \left( {\mathbf{D}} \right) = 4\pi \rho \] where \[\rho \] is the density of free charges, we can write the electric field, inside each region, as follows: \[\begin{gathered} {E_1} = \frac{{4\pi {\sigma _0}}}{{{\varepsilon _1}}}, \hfill \\ \hfill \\ {E_2} = \frac{{4\pi {\sigma _0}}}{{{\varepsilon _2}}}, \hfill \\ \hfill \\ {E_3} = \frac{{4\pi {\sigma _0}}}{{{\varepsilon _3}}}. \hfill \\ \end{gathered} \] where \[{{\sigma _0}}\] is the charge density, in absence of dielectrics |dw:1434516781566:dw|
Michele_Laino
  • Michele_Laino
now we can write: \[\begin{gathered} \Delta \varphi = \left( {{E_1} + {E_2}} \right)\frac{d}{2} = \frac{{4\pi {\sigma _0}d}}{2}\left( {\frac{1}{{{\varepsilon _1}}} + \frac{1}{{{\varepsilon _2}}}} \right) = \frac{{{q_0}}}{C} \hfill \\ \hfill \\ C = \frac{{2S}}{{4\pi d}}\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}} = 2{C_0}\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}} \hfill \\ \end{gathered} \] where C_0 is the capacitance in absence of dielectrics. Please note that I have used the CGS system
Michele_Laino
  • Michele_Laino
similarly we can write: \[C = \frac{{2S}}{{4\pi d}}\frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}} = 2{C_0}\frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}}\]
Michele_Laino
  • Michele_Laino
now if we sum both last equations side by side, we get: \[C = {C_0}\left( {\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}} + \frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}}} \right)\] finally substituting your values, we get: \[\begin{gathered} C = {C_0}\left( {\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}} + \frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}}} \right) = {C_0}\left( {\frac{{6 \times 2}}{{6 + 2}} + \frac{{6 \times 4}}{{6 + 4}}} \right) = \hfill \\ \hfill \\ = {C_0}\left( {\frac{{12}}{5} + \frac{3}{2}} \right) = 3.9{C_0} \hfill \\ \end{gathered} \]

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