<h3 id="back-link"><a href="/#6.5">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
  <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>

    <h3>Solution</h3>
    <p>
      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
      $$d\vec{F} = \vec{E}dq = \pm\frac{\sigma}{2\varepsilon_0}dq\~\hat{E}$$
      As $dq = \pm\sigma\~dS$,
      $$dF = \frac{\sigma^2\~dS}{2\varepsilon_0}$$
      and moving $dS$ to another member of equation,
      $$\boxed{P_e=\frac{dF}{dS}=\frac{\sigma^2}{2\varepsilon_0}}$$
      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>
      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
      $$\vec{E} = \vec{E_1} + \vec{E_2}$$
      $E = \frac{\sigma}{\varepsilon_0}$
      $$\boxed{\sigma = \varepsilon_0 E} \\;(1)$$
      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,
      $$P_e = \frac{dF}{dS} = \frac{\sigma E}{2} \\;(2)$$
      Putting $(1)$ into $(2)$
      $$\boxed{P_e = \frac{\varepsilon_0\~E^2}{2}}$$
      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>
      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,
      $$\vec{E} = 2\vec{E_1} = 2\vec{E_2}$$
      $$E = \frac{\sigma}{\varepsilon_0}$$
      with $\varepsilon_0$=8.85$\times$10$^{-12}$ F/m
      <div class="scroll-wrapper">
        $$\boxed{\sigma = \varepsilon_0\~E = 8.85\times10^{-6}\~\rm{C/m^2}=8.85\~\rm{\mu C/m^2}} \\;(4)$$
      </div>
      Since $P_e = \frac{\sigma^2}{2\varepsilon_0}$ and According $(4)$
      $$\boxed{P_e = \frac{\varepsilon_0\~E^2}{2}=4.425\~\rm{Pa}}$$
    </p>

    <p 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>



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