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

1. kanwal32

|dw:1434454283651:dw|

2. kanwal32

|dw:1434454418404:dw|

3. kanwal32

lower portion of the capacitance are in parralel so Ceq=$\frac{ 2EoA/2 }{ d/2 }+\frac{ 4EoA/2 }{ d/2 }$

4. kanwal32

EoA/d=C

5. kanwal32

2C+4C=Ceq

6. kanwal32

6c=Ceq

7. kanwal32

now the lower and upper one are in series

8. kanwal32

$\frac{ d/2 }{ 6kEoA }+\frac{ 1 }{ 6C }$

9. kanwal32

$\frac{ 3 }{ 12C }$=1/Ceq

10. kanwal32

Ceq=4C but my book says 3.9

11. kanwal32

3.9C

12. kanwal32

13. kanwal32

@IrishBoy123 hlp

14. 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|

15. 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

16. Michele_Laino

similarly we can write: $C = \frac{{2S}}{{4\pi d}}\frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}} = 2{C_0}\frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}}$

17. 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: $\begin{gathered} C = {C_0}\left( {\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}} + \frac{{{\varepsilon _1}{\varepsilon _3}}}{{{\varepsilon _1} + {\varepsilon _3}}}} \right) = {C_0}\left( {\frac{{6 \times 2}}{{6 + 2}} + \frac{{6 \times 4}}{{6 + 4}}} \right) = \hfill \\ \hfill \\ = {C_0}\left( {\frac{{12}}{5} + \frac{3}{2}} \right) = 3.9{C_0} \hfill \\ \end{gathered}$