Repeat Exercise 21.39, but now let the charge at the origin be -4.00nC.

Given data;

Two charges${\mathrm{q}}_1$ and${\mathrm{q}}_2$

role="math" localid="1668334639650" ${\mathrm{q}}_1={\mathrm{q}}_2=4\mathrm{nC}$

Formula;

Electric field due to charge

$\mathrm{E}=\frac{\mathrm{kq}}{{\mathrm{r}}^2}\left(\hat{\mathrm{i}}\right)$ .......... (1)
Formula of electric field in form of function

$\mathrm{E}=2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}$ .......... (2)

(a) Both charges are negative

At point A

Electric field due to${\mathrm{q}}_1$ charge

${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right)$

Electric field due to${\mathrm{q}}_2$ charge

${\mathrm{E}}_2=\frac{{\mathrm{kq}}_2}{{\mathrm{r}}^2}\left(\hat{\mathrm{i}}\right)$

Net electric field

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1\left(-\hat{\mathrm{i}}\right)+{\mathrm{E}}_2\left(\hat{\mathrm{i}}\right) \\ =0 \end{gathered}$$

At point B

the angle between field line and the horizontal to be$\theta$, thehorizontalcomponentof first charge and the horizontal of second charge are equal inmagnitudebut opposite indirectionso the net field hasn't a horizontal component, the vertical component of first charge and second charge are equal in magnitude and in the same direction so the net field has only a vertical component in $-\hat{\mathrm{j}}$.

$\mathrm{E}=-2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{-2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\hat{\mathrm{j}}$

At point

the angle between field line and the horizontal to be, the horizontal component of first charge and the horizontal of second charge are equal in magnitude but opposite in direction so the net field hasn't a horizontal component, the vertical component of first charge and second charge are equal in magnitude and in the same direction so the net field has only a vertical component in$\hat{\mathrm{j}}$

$\mathrm{E}=2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\hat{\mathrm{j}}$

Hence, the direction of the net electric field due to two charges with negative charge at points$\mathrm{A}:\mathrm{E}=0,\mathrm{B}:-\hat{\mathrm{j}},\mathrm{C}:\hat{\mathrm{j}}$

(b)${\mathrm{q}}_1$ is positive and${\mathrm{q}}_2$ is negative charge

At point A

Electric field due to${\mathrm{q}}_1$ charge

role="math" localid="1668336623669" ${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right)$

Electric field due to${\mathrm{q}}_2$ charge

${\mathrm{E}}_2=\frac{{\mathrm{kq}}_2}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right)$

Net electric field

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1\left(-\hat{\mathrm{i}}\right)+{\mathrm{E}}_2\left(-\hat{\mathrm{i}}\right) \\ =\frac{2\mathrm{Kq}}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right) \end{gathered}$$

At point B

the angle between field line and the horizontal to be $\theta$, the horizontal component of first charge and the horizontal of second charge are equal in magnitude and in the same direction so the net field has a horizontal component in direction of $-\hat{\mathrm{i}}$, the vertical component of first charge and second charge are equal in magnitude but opposite in direction so the net field hasn't a vertical component.

$\mathrm{E}=2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\left(-\hat{\mathrm{i}}\right)$

At point C

the angle between field line and the horizontal to be, the horizontal component of first charge and the horizontal of second charge are equal in magnitude and in the same direction so the net field has a horizontal component in direction of $-\hat{\mathrm{i}}$, the vertical component of first charge and second charge are equal in magnitude but opposite in direction so the net field hasn't a vertical component.

role="math" localid="1668336352583" $\mathrm{E}=2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\left(-\hat{\mathrm{i}}\right)$

Hence, the direction of the net electric field due to two charges with one is positive and second is negative charge at points$\mathrm{A}:-\hat{\mathrm{i}},\mathrm{B}:-\hat{\mathrm{i}},\mathrm{C}:-\hat{\mathrm{i}}$

(a) Both charges are negative

At point A

Electric field due to${\mathrm{q}}_1$ charge

${\mathrm{E}}_1=\frac{{\mathrm{kq}}_1}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right)$

Electric field due to${\mathrm{q}}_2$ charge

${\mathrm{E}}_2=\frac{{\mathrm{Kq}}_2}{{\mathrm{r}}^2}\left(-\hat{\mathrm{i}}\right)$

Net electric field

$$\begin{gathered} \mathrm{E}={\mathrm{E}}_1\left(-\hat{\mathrm{i}}\right)+{\mathrm{E}}_2\left(\hat{\mathrm{i}}\right) \\ =0 \end{gathered}$$

At point

the angle between field line and the horizontal to be $\theta$, thehorizontalcomponentof first charge and the horizontal of second charge are equal inmagnitudebut opposite indirectionso the net field hasn't a horizontal component, the vertical component of first charge and second charge are equal in magnitude and in the same direction so the net field has only a vertical component in $-\hat{\mathrm{j}}$.

$\mathrm{E}=-2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{-2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\hat{\mathrm{j}}$

At point C

the angle between field line and the horizontal to be $\theta$, the horizontal component of first charge and the horizontal of second charge are equal in magnitude but opposite in direction so the net field hasn't a horizontal component, the vertical component of first charge and second charge are equal in magnitude and in the same direction so the net field has only a vertical component in$\hat{\mathrm{j}}$

role="math" localid="1668336444858" $\mathrm{E}=2\mathrm{Esin}\left(\mathrm{\theta }\right)\hat{\mathrm{j}}=\frac{2{\mathrm{Kq}}_1\mathrm{y}}{{\mathrm{r}}^3}\hat{\mathrm{j}}$

Hence, the direction of the net electric field due to two charges with negative charge at points$\mathrm{A}:\mathrm{E}=0,\mathrm{B}:-\hat{\mathrm{j}},\mathrm{C}:\hat{\mathrm{j}}$