91影视

Given data;

Two charges $q_{1}and{q}_2$

$q_1=q_2$

Formula;

Electric field due to charge

$\mathit{E}\mathbf{=}\frac{\mathbf{K}\mathbf{q}}{{\mathbf{r}}^{\mathbf{2}}}\widehat{\mathbf{\left(}\mathbf{i}\mathbf{\right)}}$ .......... (1)
Formula of electric field in form of function

$\mathit{E}\mathbf{=}\mathbf{2}\mathit{E}\mathit{s}\mathit{i}\mathit{n}\mathbf{\left(}\mathit{胃}\mathbf{\right)}\hat{\mathbf{j}}$ .......... (2)

(a) Both charges are negative

At point

Electric field due to q₁ charge

$E_1=\frac{K{q}_1}{r^2}\widehat{\left(i\right)}$

Electric field due to q₂ charge

$E_2=\frac{K{q}_2}{r^2}\widehat{\left(i\right)}$

Net electric field

$E=E_1\widehat{(-i)}+E_2\left(\hat{i}\right)$

At point

Theanglebetween field line and the horizontal to be, 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$-\left(\hat{j}\right)$.

$E=-2E\sin \left(胃\right)\left(\hat{j}\right)=\frac{{\displaystyle -2K{q}_{1}y}}{{\displaystyle r^3}}\left(\hat{j}\right)$

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 $\left(\hat{j}\right)$

$E=-2E\sin \left(胃\right)\left(\hat{j}\right)=\frac{{\displaystyle 2K{q}_{1}y}}{{\displaystyle r^3}}\left(\hat{j}\right)$

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

(b) Both charges are positive

At point A

Electric field due to charge

$E_1=\frac{K{q}_1}{r^2}\widehat{\left(i\right)}$

Electric field due to charge

$E_2=\frac{K{q}_2}{r^2}\widehat{(-i)}$

Net electric field

$$\begin{gathered} E=E_1\widehat{(-i)}+E_2\left(\hat{i}\right) \\ =0 \end{gathered}$$

At point

The angle between field line and the horizontal to be$胃$, 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 .

$E=2E\sin \left(胃\right)\left(\hat{j}\right)=\frac{{\displaystyle 2K{q}_{1}y}}{{\displaystyle r^3}}\left(\hat{j}\right)$

At point

The angle between field line and the horizontal to be$胃$, 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 .

$E=-2E\sin \left(胃\right)\left(\hat{j}\right)=\frac{{\displaystyle -2K{q}_{1}y}}{{\displaystyle r^3}}\left(\hat{j}\right)$

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

(c) q₁ is positive and q₂ is negative charge

At point

Electric field due to q1 charge

$E_1=\frac{K{q}_1}{r^2}\widehat{\left(i\right)}$

Electric field due to q2 charge

$E_2=\frac{K{q}_2}{r^2}\widehat{\left(i\right)}$

Net electric field

$$\begin{gathered} E=E_1\widehat{(-i)}+E_2\left(\hat{i}\right) \\ =\frac{2K{q}_1}{r^2}\widehat{\left(i\right)} \end{gathered}$$

At point

The angle between field line and the horizontal to be$胃$, 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 , 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.

$E=2E\sin \left(胃\right)\hat{j}=\frac{2K{q}_{1}y}{r^3}\hat{i}$

At point

the angle between field line and the horizontal to be$胃$, 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, 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.

$E=2E\sin \left(胃\right)\hat{j}=\frac{2K{q}_{1}y}{r^3}\hat{i}$

Hence, the direction of the net electric field due to two charges with one is positive and second is negative charge at points$A:E=0,B:\hat{J},C-\hat{J}$