Question

Two point particles, one with charge +8.00 nC and the other with charge -2.00 nC, are separated by 4.00 m. The electric field in N/C midway between them is:

Answer 1

The electric field at the middle of a certain distance (d=4 m) can be found as:

$\overrightarrow{E_c}=\overrightarrow{E_{1c}}+\overrightarrow{E_{2c}}=k\dfrac{q_1}{{r_{1c}}^2}\,\widehat{r_{1c}}+k\dfrac{q_2}{{r_{2c}}^2}\,\widehat{r_{2c}}$

Following this we have to set the coordinates for the system so that r₁=(0,0)m and r₂=(4,0)m are the position vectors for the charges q₁= +8 nC and q₂= -2 nC and then we find everything related to the center of the distance r_c=(d/2,0)=(2,0)m:

$\overrightarrow{r_{1c}}=\overrightarrow{r_c}-\overrightarrow{r_1}=(2,0)m-(0,0)m=(2,0)m \\ \overrightarrow{r_{2c}}=\overrightarrow{r_c}-\overrightarrow{r_2}=(2,0)m-(4,0)m=(-2,0)m \\ r_{1c}=r_{2c}=d/2=2\,m \\ \widehat{r_{1c}}=\cfrac{\overrightarrow{r_{1c}}}{r_{1c}}=\frac{(2,0)}{2}=(1,0) \\ \widehat{r_{2c}}=\cfrac{\overrightarrow{r_{2c}}}{r_{2c}}=\frac{(-2,0)}{2}=(-1,0)$

With this information we substitute to find $\overrightarrow{E_c}$:

$\overrightarrow{E_c}=k\dfrac{q_1}{{r_{1c}}^2}\,\widehat{r_{1c}}+k\dfrac{q_2}{{r_{2c}}^2}\,\widehat{r_{2c}}$

$\overrightarrow{E_c}=(8.988\times10^9\frac{Nm^2}{C^2})\frac{(8\times10^{-9}C)}{(2\,m)^2}(1,0)+(8.988\times10^9\frac{Nm^2}{C^2})\frac{(-2\times10^{-9}C)}{(2\,m)^2}(-1,0)$

$\overrightarrow{E_c}=(17.976)(1,0)\frac{N}{C}+(-4.494)(-1,0) \frac{N}{C}=(17.976,0)\frac{N}{C}+(4.494,0)\frac{N}{C}$

$\to \overrightarrow{E_c}=(22.47,0)\frac{N}{C} \implies {E_c}=22.47 \,\frac{N}{C}$

In conclusion, the electric field midway between the charges is E_c=22.47 N/C.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.