Question

A$10-\mathrm{cm}$long thin glass rod uniformly charged to$+10\mathrm{nC}$and a $10-\mathrm{cm}$-long thin plastic rod uniformly charged to$-10\mathrm{nC}$are placed side by side, $4.0\mathrm{cm}$apart. What are the electric field strengths${\mathrm{E}}_1$to${\mathrm{E}}_3$at distances$1.0\mathrm{cm},2.0\mathrm{cm}$, and from the glass rod a$3.0\mathrm{cm}$long the line connecting the midpoints of the two rods?

Short answer

Electric field at distance $3\mathrm{cm}$is$22.84\times {10}^4\mathrm{N}/\mathrm{C}$.

Electric field at distance $2\mathrm{cm}$is $16.7\times {10}^4\mathrm{N}/\mathrm{C}$.

Electric field at distance$1\mathrm{cm}$is$22.84\times {10}^4\mathrm{N}/\mathrm{C}$.

Step-by-step solution

Figure for two rods with electric field strength

Figure is,

Calculation for Electric field strength at distance r=1 cm

The electric field produced by a rod at a distance perpendicular to its midpoint is calculated as follows:

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

$\mathrm{k}=9\times {10}^9\mathrm{N}·{\mathrm{m}}^2/{\mathrm{C}}^2$.

$\mathrm{q}$-charge

The electric field at $\mathrm{r}=1\mathrm{cm}$:

An electric field is produced by both glass and plastic rods, and the direction of the field is determined by the charge on the rods. which is in the$(+)\mathrm{x}$direction.

We'll also include both electric fields because both rods are pointed in the same direction.

Electric field is,

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

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

$=(17.7+5.14)\times {10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_1=22.84\times {10}^4\mathrm{N}/\mathrm{C}$

Calculation of electric field at distance r=2.0 cm and r=3.0 cm

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

${\mathrm{E}}_2={\mathrm{E}}_{\text{glass}}+{\mathrm{E}}_{\text{plastic}}$

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

$=(8.35+8.35)\times {10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_2=16.7\times {10}^4\mathrm{N}/\mathrm{C}$

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

${\mathrm{E}}_3={\mathrm{E}}_{\text{glass}}+{\mathrm{E}}_{\text{plastic}}$

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

$=(5.14+17.7)\times {10}^4\mathrm{N}/\mathrm{C}$

${\mathrm{E}}_3=22.84\times {10}^4\mathrm{N}/\mathrm{C}$