Question

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Answer 1

Treat it as two capacitors in series.
\[\frac{ 1 }{ C }=\frac{ 1 }{ C_1 }+\frac{ 1 }{ C_2 }\]
where
\[C_1=\frac{2 \kappa_1\epsilon_0A }{ d }\]
and
\[C_2=\frac{2 \kappa_2\epsilon_0A }{ d }\]

Answer 2

\[C=\frac{ C_1C_2 }{ C_1+C_2 }\]
\[C=\frac{ \frac{ 4\kappa_1\kappa_2\epsilon_0^2A^2 }{ d^2 } }{\frac{ 2\epsilon_0A(\kappa_1+\kappa_2) }{ d } }\]
\[C=2\frac{ \epsilon_0A }{ d }\frac{ \kappa_1\kappa_2 }{ \kappa_1+\kappa_2 }\]

So basically you can combine the dielectrics in series and multiply by twice the plate's capacitance with no dielectric

Answer 3

here we have to note that we can apply this equation of Maxwell:
\[\Large div{\mathbf{D}} = 4\pi \rho \]
where \(\textbf{D}\) is the electrostic induction, and \(\rho\) is the charge density of \(free\) electrical charge, namely it is the non-polarization electrical charge. So we get:
\[\Large {\varepsilon _1}{E_1} = 4\pi \sigma \]
wherein:
\[\Large Q = \sigma A\]
\(Q\) is the magnitude of electrical charge on each pkate of the capacitor, and \(A\) is the corresponding surface.
Next, at the separation surface of the two dielectric mediums we have:
\[\Large {D_{{n_1}}} = {D_{{n_2}}}\]
namely, the normal component of the electrical induction has to be conseved, so we can write:
\[\Large {\varepsilon _1}{E_1} = {\varepsilon _2}{E_2} \Rightarrow {E_2} = \frac{{4\pi \sigma }}{{{\varepsilon _2}}}\]
being:
\[\Large {\varepsilon _1}{E_1} = 4\pi \sigma \]
Subsequently I apply the subsequent Maxwell's equation:
\[\Large rot{\mathbf{E}} = {\mathbf{0}} \Rightarrow {E_1}{d_1} + {E_2}{d_2} = V\]
and after a substitution, I get:
\[\Large 4\pi \frac{{{d_2}{\varepsilon _1} + {d_1}{\varepsilon _2}}}{{{\varepsilon _1}{\varepsilon _2}}} = \frac{A}{C}\]
from which I get:
\[\Large C = \frac{{A{\varepsilon _1}{\varepsilon _2}}}{{4\pi \left( {{d_2}{\varepsilon _1} + {d_1}{\varepsilon _2}} \right)}}\]
In my computaton I have used the \(CGS\) system of units of measures

Answer 4

here \(V\) is the voltage across the capacitor

Answer 5

furthermore, we have: \(d1+d2=d\), so, if \(d1=d2=d/2\), the requested capacity, is:
\[\Large C = \frac{{A{\varepsilon _1}{\varepsilon _2}}}{{2\pi d\left( {{\varepsilon _1} + {\varepsilon _2}} \right)}} = 2{C_0}\frac{{{\varepsilon _1}{\varepsilon _2}}}{{{\varepsilon _1} + {\varepsilon _2}}}\]
wherein \(C_0\) is the capacity of a capacitor with no medium enclosed between its plates, and with the same geometrical characteristics of the present capacitor. Finally, the dielectric constants, are indicated with symbols \(\varepsilon_1,\; \varepsilon_2\)

Answer 6

Got it! Thank you both :)

If I interpret correctly, you're working the capacitance in below sequence of steps :

1) find the magnitude of electric field in both dielectrics using gauss/maxwell's eqn
2) find the potential difference across each dielectric
3) C = Q/V

Answer 7

1) electric field
\(E_1 = \dfrac{E_0}{k_1}\)
\(E_2 = \dfrac{E_0}{k_2}\)

2) potential difference
\(V = E_1\dfrac{d}{2} + E_2\dfrac{d}{2} = \dfrac{E_0d}{2}(\dfrac{1}{k_1}+\dfrac{1}{k_2}) = \dfrac{Qd}{2\epsilon_0 A}(\dfrac{1}{k_1}+\dfrac{1}{k_2}) \)

3) capacitance
\(C = \dfrac{Q}{V} = \dfrac{2\epsilon_0 A}{d}(\dfrac{1}{k_1}+\dfrac{1}{k_2})^{-1} = 2C_0\dfrac{k_1k_2}{k_1+k_2} \)

Thank you so much for the general method !

Answer 8

please note that, if we consider a little cylinder with base area \(dS\) (it is a differential\) acreoss the separation surface of the two mediums, and we apply the theorem of Gauss in this form, we get:

\[\Large \nabla \cdot {\mathbf{D}} = 0 \Rightarrow {\varepsilon _1}{E_1}dS - {\varepsilon _2}{E_2}dS = 0\]
and therefore:
\[\Large {\varepsilon _1}{E_1} = {\varepsilon _2}{E_2}\]