Two very large parallel sheets are 5.00 cm apart. Sheet A carries a uniform surface charge density of -8.80 micro coulomb per meter square, and sheet B, which is to the right of A, carries a uniform charge density of +11. micro coulomb per meter square. Assume that the sheets are large enough to be treated as infinite. Find the magnitude and direction of the net electric field these sheets produce at a point

(a) 4.00 cm to the right of sheet A.

(b) 4.00 cm to the left of sheet A.

(c) 4.00 cm to the right of sheet B.

$$\begin{gathered} {\sigma }_A=-8.80\mu C/m^2=-8.80\times {10}^{-6}C,{\sigma }_B=+11.6\mu C/m^2=+11.6\times {10}^{-6}C, \\ {\epsilon }_o=8.85\times {10}^{-12}{C}^2/N.m^2,\mathrm{d}=5.0\mathrm{cm} \end{gathered}$$

We know that each sheet produces an electric field and we also know that the infinity sheets produce a field that is Independent of distance from the sheet and its field is given by

$E=\frac{\sigma }{2{\epsilon }_o}$

For the first point, (which is presented in the figure below and we also found the distance between sheet B and the given point as you see below), we need to find the vector sum of these two electric fields. Noting that the direction of the electric field of each sheet is represented in the second figure below. We choose the right direction to be our positive x-direction.

Hence,

$$\begin{gathered} E_{net}=-E_A-E_B \\ {E}_{net}=-\frac{\left|{\sigma }_A\right|}{2{\epsilon }_o}-\frac{\left|{\sigma }_B\right|}{2{\epsilon }_o} \\ {E}_{net}=\left(\frac{\left|{\sigma }_A\right|-\left|{\sigma }_B\right|}{2{\epsilon }_o}\right) \end{gathered}$$

Put the given from above

b) By the same approach as part (a), and from the second figure below, the net electric field of this position is given by

$$\begin{gathered} E_{net}=E_A-E_B \\ {E}_{net}=\frac{\left|{\sigma }_A\right|}{2{\epsilon }_o}-\frac{\left|{\sigma }_B\right|}{2{\epsilon }_o} \\ {E}_{net}=\left(\frac{\left|{\sigma }_A\right|-\left|{\sigma }_B\right|}{2{\epsilon }_o}\right) \end{gathered}$$

Put the given from above

$$\begin{gathered} E_{net}=\left(\frac{\left|-8.80\times {10}^{-6}\right|-\left|+11.6\times {10}^{-6}\right|}{2\times 8.85\times {10}^{-12}}\right) \\ {E}_{net}=-\left(1.58\times {10}^{5}N/C\right)i \end{gathered}$$

$$\begin{gathered} E_{net}=E_B-E_A \\ {E}_{net}=\frac{\left|{\sigma }_B\right|}{2{\epsilon }_o}-\frac{\left|{\sigma }_A\right|}{2{\epsilon }_o} \\ {E}_{net}=\frac{\left|{\sigma }_B\right|-\left|{\sigma }_A\right|}{2{\epsilon }_o} \end{gathered}$$

Put the given from above

$$\begin{gathered} E_{net}=\frac{\left|-11.6\times {10}^{-6}\right|-\left|-8.80\times {10}^{-6}\right|}{2\times 8.85\times {10}^{-12}} \\ {E}_{net}=\left(1.58\times {10}^{6}N/C\right)i \end{gathered}$$