Updated greek laters @ latex compiling

astrosander edited
revision #10185 parent #9122 GitHub dbb31ba ← older newer →
@@ -83,11 +83,11 @@
According to superposition principle between plates, $\vec{E_1}+\vec{E_2} = \vec{0}$, hence, $\vec{E_1}=-\vec{E_2}$ or $E_1 = E_2 = \frac{\sigma}{2\varepsilon_0}$. So, plates must have charges of same sign (+) or (-). The outside field is
$$\vec{E} = \vec{E_1} + \vec{E_2}$$
$E = \frac{\sigma}{\varepsilon_0}$
− $$\boxed{\sigma = \varepsilon_0 E} \;(1)$$
+ $$\boxed{\sigma = \varepsilon_0 E} \quad(1)$$
The force that acts over an element of charge $dq$ in a plate is $dF_i = E_i~dq$ (i =1,2), but $E_i = \frac{E}{2}$ so,
$$dF = \frac{E}{2}dq = \frac{E}{2}\sigma~dS$$
Hence,
− $$P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \;(2)$$
+ $$P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \quad(2)$$
Putting $(1)$ into $(2)$
$$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}}$$
c) See following figure
@@ -107,7 +107,7 @@
$$E = \frac{\sigma}{\varepsilon_0}$$
with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m
<div class="scroll-wrapper">
− $$\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \;(4)$$
+ $$\boxed{\sigma = \varepsilon_0~E = 8.85\times10^{-6}~\rm{C/m^2}=8.85~\rm{\mu C/m^2}} \quad(4)$$
</div>
Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$
$$\boxed{P_e = \frac{\varepsilon_0~E^2}{2}=4.425~\rm{Pa}}$$
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