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  <h4>Statement</h4>
  <p>
    $6.5.1$
    a) The surface densities of charges in two planes are $\pm \sigma$. The distance between them is much less than the dimensions of them. What force per unit of surface of plane (electric pressure) will act? 
  </p>
  <p class="TxtSolutions">
    b) The intensity of electric field between two parallel planes is null, out of them is $E$. Find the surface density of charge in the planes. What electric pressure acts over the planes?
  </p>
  <p class="TxtSolutions">
    c) The field intensity between two parallel planes is 10 000 V/cm, while out of them is null. Find the electric pressure over each plane and the surface density of charge.
  </p>
  <h4>Solution</h4>
  <p class="TxtSolutions">
    a) The absolute value of the electric field created by each plate is $E = \frac{\sigma}{2\varepsilon_0}$, between them is $2E = \frac{\sigma}{\varepsilon_0}$ (superposition). The force that acts over a differential of charge $dq$ is
  </p>
  <center>
    $d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq\~\hat{E}$
  </center>
  <p calss="TxtSolutions">
    As $dq = \pm\sigma\~dS$,
  </p>
  <center>
    $dF = \frac{\sigma^2\~dS}{2\varepsilon_0}$
  </center>
  <p class="TxtSolutions">
    and moving $dS$ to another member of equation,
  </p>
  <center>
    $\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$
  </center>
  <p class="TxtSolutions">
    b) Let's see following figure
  </p>
  <center>
    <figure>
      <img src="drawb.png"
        loading="lazy" alt="6.5.1" width="300" />
      <figcaption>
        Electric field of plates
      </figcaption>
    </figure>
  </center>
  <p class="TxtSolutions">
    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
  </p>
  <center>
    $\vec{E} = \vec{E_1} + \vec{E_2}$
  </center>
  <center>
    $E = \frac{\sigma}{\varepsilon_0}$
  </center>
  <center>
    $\boxed{\sigma = \varepsilon_0 E} \\;(1)$
  </center>
  <p class="TxtSolutions">
    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,
  </p>
  <center>
    $dF = \frac{E}{2}dq = \frac{E}{2}\sigma\~dS$
  </center>
  <p class=" TxtSolutions">
    Hence,
  </p>
  <center>
    $P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \\;(2)$
  </center> 
  <p class="TxtSolutions">
    Putting $(1)$ into $(2)$
  </p>
  <center>
    $\boxed{P_e = \frac{\varepsilon_0\~E^2}{2}}$
  </center>
  <p class="TxtSolutions">
    c) See following figure
  </p>
  <center>
    <figure>
      <img src="drawc.png"
        loading="lazy" alt="6.5.1" width="300" />
      <figcaption>
        Electric field of plates
      </figcaption>
    </figure>
  </center>
  <p class="TxtSolutions">
    Superposition of electric fields must be zero outside of capacitor, so $\vec{E_1} = -\vec{E_2}$, so plates must have opposed signs. Inside is valid that $\vec{E_1} = \vec{E_2}$ so,
  </p>
  <center>
    $\vec{E} = 2\vec{E_1} = 2\vec{E_2}$
  </center>
  <center>
    $E = \frac{\sigma}{\varepsilon_0}$
  </center>
  <p class="TxtSolutions">
    with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m
  </p>
  <center>
    $\boxed{\sigma = \varepsilon_0\~E = 8.85\times10^{-6}\~\rm{C/m^2}=8.85\~\rm{\mu C/m^2}} \\;(4)$
  </center>
  <p class="TxtSolutions">
    Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$
  </p>
  <center>
    $\boxed{P_e = \frac{\varepsilon_0\~E^2}{2}=4.425\~\rm{Pa}}$
  </center>
    <p class="TxtSolutions" style="text-align: right; font-style: italic; font-size: 14;">   
        BSc. Luis Daniel Fernández Quintana<br>
        Physics Department (FCNE)<br>
        Universidad de Oriente, Cuba<br>
    </p>
</article>