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kanwal32
 one year ago
Equivalent capacitance dielectric (if possible tell me what am i doing wrong
kanwal32
 one year ago
Equivalent capacitance dielectric (if possible tell me what am i doing wrong

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kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434454283651:dw

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0dw:1434454418404:dw

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0lower portion of the capacitance are in parralel so Ceq=\[\frac{ 2EoA/2 }{ d/2 }+\frac{ 4EoA/2 }{ d/2 }\]

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0now the lower and upper one are in series

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ d/2 }{ 6kEoA }+\frac{ 1 }{ 6C }\]

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ 3 }{ 12C }\]=1/Ceq

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0Ceq=4C but my book says 3.9

kanwal32
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 @radar @Michele_Laino

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1using 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
 one year ago
Best ResponseYou've already chosen the best response.1now 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
 one year ago
Best ResponseYou've already chosen the best response.1similarly 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
 one year ago
Best ResponseYou've already chosen the best response.1now 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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