91Ó°ÊÓ

Figure is,

The following formula is used to compute the electric field produced by a rod at a distance perpendicular to its midpoint:

$\mathrm{E}=\frac{\mathrm{kq}}{\mathrm{r}\sqrt{{\mathrm{r}}^2+\frac{{\mathrm{l}}^2}{4}}}$

Both glass and plastic rods will generate an electric field, and the direction of the field will be determined by the charge on the rods.which will be in thelocalid="1649133464706" $-\mathrm{X}$direction.

For $\mathrm{r}=1.0\mathrm{cm}$

${\mathrm{E}}_1={\mathrm{E}}_{\text{glass}}-{\mathrm{E}}_{\text{plastic}}$

${\mathrm{E}}_1=\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{(0.01\mathrm{m})\left(\sqrt{(0.01\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right.}\right)-\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{\left(\right(0.04-0.01\left)\mathrm{m}\right)\left(\sqrt{\left(\right(0.04-0.01)\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right.}\right)$

${\mathrm{E}}_1=(17.7-5.14)×{10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_1=12.56×{10}^4\mathrm{N}/\mathrm{C}$

For $\mathrm{r}=2.0\mathrm{cm}$:

localid="1649132926806" ${\mathrm{E}}_2={\mathrm{E}}_{\text{glass}}-{\mathrm{E}}_{\text{plastic}}$

${\mathrm{E}}_2=\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{(0.02\mathrm{m})\left(\sqrt{(0.02\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right.}\right)-\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{\left(\right(0.04-0.02\left)\mathrm{m}\right)\left(\sqrt{\left(\right(0.04-0.02)\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right)}\right)$

${\mathrm{E}}_2=(8.35-8.35)×{10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_2=0\mathrm{N}/\mathrm{C}$

For $\mathrm{r}=3.0\mathrm{cm}$

localid="1649132933182" ${\mathrm{E}}_3={\mathrm{E}}_{\text{glass}}-{\mathrm{E}}_{\text{plastic}}$

${\mathrm{E}}_3=\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{(0.03\mathrm{m})\left(\sqrt{(0.03\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right.}\right)-\left(\frac{\left(9×{10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2\right)\left(10×{10}^{-9}\mathrm{C}\right)}{\left(\right(0.04-0.03\left)\mathrm{m}\right)\left(\sqrt{\left(\right(0.04-0.03)\mathrm{m}{)}^2+{\left(\frac{0.1}{2}\mathrm{m}\right)}^2}\right)}\right)$

${\mathrm{E}}_3=(5.14-17.7)×{10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_3=-12.56\mathrm{N}/\mathrm{C}$