Question

You've hung two very large sheets of plastic facing each other with distance$\mathrm{d}$between them, as shown in$\mathrm{FIGURE}\mathrm{EX}23.20$. By rubbing them with wool and silk, you've managed to give one sheet a uniform surface charge density and the other a uni${\mathrm{\eta }}_1=-{\mathrm{\eta }}_0$form surface charge density ${\mathrm{\eta }}_2=+3{\mathrm{\eta }}_0$. What are the electric field vectors at points$1$,$2$and$3$$?$

Short answer

Electric field at point$1$is$\frac{{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$,at point $2$is $\frac{2{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$,at point $3$is $\frac{{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$.

Step-by-step solution

Figure for electric field at point1,2and3in sheet

Direction of$\mathrm{E}$of$1$and $2$away from${\mathrm{\eta }}_{\mathrm{e}}$charge sheet and direction of $\mathrm{E}$at$3$,away from$+{\mathrm{\eta }}_{\mathrm{e}}$charge sheet.

Calculation of electric field at point1

${\mathrm{E}}_{+}$due to $3{\mathrm{\eta }}_0$

${\mathrm{E}}_{-}$due to $-{\mathrm{\eta }}_0$

Electric field at $1$:

$\mathrm{E}={\mathrm{E}}_{+}-{\mathrm{E}}_{-}$

$=\frac{3{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}-\frac{{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}$

$\mathrm{E}=\frac{{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$

Calculation of electric field at point2and3

Electric field at $2$:

$\mathrm{E}={\mathrm{E}}_{+}+{\mathrm{E}}_{-}$

$\mathrm{E}=\frac{3{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}+\frac{{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}$

$\mathrm{E}=\frac{2{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$

Electric field at $3$:

$\mathrm{E}={\mathrm{E}}_{+}+{\mathrm{E}}_{-}$

$=\frac{3{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}-\frac{{\mathrm{\eta }}_0}{2{\mathrm{\eta }}_0}$

$\mathrm{E}=\frac{{\mathrm{\eta }}_0}{{\mathrm{\eta }}_0}$