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I'm still not confident with these kinds of problems. Hopefully I got it right, but can someone double check my work? Thanks!
Two parallel, infinite planes are separated by a distance d. Find the electric field everywhere (a) if both planes carry a surface charge density \sigma and (b) if one plane has a charge density \sigma and the other - \sigma.
Here's my attempt. I'm only trying part (a) here.
The electric field is perpendicular to the planes because of symmetry. The components of the electric field parallel to the planes add to zero because of symmetry. Therefore, the electric field is perpendicular to the planes. So we can place a cylinder anywhere on the plane and use Gauss’ Law to compute the flux passing through the cylinder. The flux passing through the top of the cylinder will be half of the total flux.
<br /> \begin{array}{l}<br /> \Phi _{total} = \Phi _{side} + \Phi _{top} + \Phi _{bottom} \\ <br /> \\ <br /> \Phi _{side} = 0{\rm{ because }}\overrightarrow E \bot {\rm{\hat n}} \\ <br /> \\ <br /> \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\,{\rm{\hat k}} \cdot {\rm{\hat n = 1 because the normal vector, \hat n, is in the direction \hat k}}{\rm{.}} \\ <br /> \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot A \\ <br /> \\ <br /> \Phi _{{\rm{bottom}}} = - \overrightarrow {E_z } \cdot - {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\, - {\rm{\hat k}} \cdot {\rm{\hat n = }} - {\rm{1 because the normal vector, \hat n, is in direction }} - {\rm{\hat k}}{\rm{.}} \\ <br /> \Phi _{{\rm{bottom}}} = \overrightarrow {E_z } \cdot A \\ <br /> \\ <br /> \Phi _{total} = 2\overrightarrow {E_z } \cdot A = \frac{{Q_{enclosed} }}{{\varepsilon _0 }},\,{\rm{ }}Q_{enclosed} = \sigma A \\ <br /> \overrightarrow {E_z } \cdot = \frac{{\sigma }}{{2\varepsilon _0 }} \\ <br /> \overrightarrow {E_z } = \frac{\sigma }{{2\varepsilon _0 }} \\ <br /> \end{array}<br />
The electric field coming out of the top of the cylinder is half of the electric field coming out of the total cylinder. Therefore, the electric field above the top infinite plane is
<br /> \overrightarrow {E_z } = \frac{{\frac{\sigma }{{2\varepsilon _0 }}}}{2} = <br /> {\frac{\sigma }{{4\varepsilon _0 }}{\rm{,\hat k}}}<br />
Similarly, the electric field below the bottom infinite plane is
<br /> {\frac{\sigma }{{4\varepsilon _0 }}, - {\rm{\hat k}}}<br />
Between the planes, the electric field lines will cancel. Therefore there are no electric field lines between the planes.
Two parallel, infinite planes are separated by a distance d. Find the electric field everywhere (a) if both planes carry a surface charge density \sigma and (b) if one plane has a charge density \sigma and the other - \sigma.
Here's my attempt. I'm only trying part (a) here.
The electric field is perpendicular to the planes because of symmetry. The components of the electric field parallel to the planes add to zero because of symmetry. Therefore, the electric field is perpendicular to the planes. So we can place a cylinder anywhere on the plane and use Gauss’ Law to compute the flux passing through the cylinder. The flux passing through the top of the cylinder will be half of the total flux.
<br /> \begin{array}{l}<br /> \Phi _{total} = \Phi _{side} + \Phi _{top} + \Phi _{bottom} \\ <br /> \\ <br /> \Phi _{side} = 0{\rm{ because }}\overrightarrow E \bot {\rm{\hat n}} \\ <br /> \\ <br /> \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\,{\rm{\hat k}} \cdot {\rm{\hat n = 1 because the normal vector, \hat n, is in the direction \hat k}}{\rm{.}} \\ <br /> \Phi _{{\rm{top}}} = \overrightarrow {E_z } \cdot A \\ <br /> \\ <br /> \Phi _{{\rm{bottom}}} = - \overrightarrow {E_z } \cdot - {\rm{\hat k}} \cdot {\rm{\hat n}} \cdot A,\,\,\,\, - {\rm{\hat k}} \cdot {\rm{\hat n = }} - {\rm{1 because the normal vector, \hat n, is in direction }} - {\rm{\hat k}}{\rm{.}} \\ <br /> \Phi _{{\rm{bottom}}} = \overrightarrow {E_z } \cdot A \\ <br /> \\ <br /> \Phi _{total} = 2\overrightarrow {E_z } \cdot A = \frac{{Q_{enclosed} }}{{\varepsilon _0 }},\,{\rm{ }}Q_{enclosed} = \sigma A \\ <br /> \overrightarrow {E_z } \cdot = \frac{{\sigma }}{{2\varepsilon _0 }} \\ <br /> \overrightarrow {E_z } = \frac{\sigma }{{2\varepsilon _0 }} \\ <br /> \end{array}<br />
The electric field coming out of the top of the cylinder is half of the electric field coming out of the total cylinder. Therefore, the electric field above the top infinite plane is
<br /> \overrightarrow {E_z } = \frac{{\frac{\sigma }{{2\varepsilon _0 }}}}{2} = <br /> {\frac{\sigma }{{4\varepsilon _0 }}{\rm{,\hat k}}}<br />
Similarly, the electric field below the bottom infinite plane is
<br /> {\frac{\sigma }{{4\varepsilon _0 }}, - {\rm{\hat k}}}<br />
Between the planes, the electric field lines will cancel. Therefore there are no electric field lines between the planes.
