91Ó°ÊÓ

From the figure we have seen that the electric field lines are coming out from the two outside charges. The upper one and the lower one. Which means that two charges are positive.

The electric field lines are also going toward the middle charge. Indicating the middle one is a negative charge. Hence, we can conclude that the charges of the three charges are: Positive, negative, positive.

It is known that the three charges are having the same magnitude. Hence, we can assume that the distance between the charges is same.

Choosing any arbitrary point to the left of the middle charge which we will analyse the net electric field. As we assume that the distance between the charges is equal therefore the angle of the first electric field, made with the horizontal line passing through the middle charge. Which is equal to the angle made by the third charge with the same line:

${\mathrm{θ}}_1={\mathrm{θ}}_2=0$

Therefore:

To analyse the electric field at the chosen point into their x-y components, the magnitude of the electric field is given by:

$\mathrm{E}=\frac{1}{4{\mathrm{Ï€Î\mu Ï\mu }}_0}×\frac{\mathrm{q}}{{\mathrm{r}}^2}$

Now, as both the electric field of E_1 and E_2 is at as the same point as from our chosen point. As their charges are having the same magnitude hence:

${\mathrm{E}}_1={\mathrm{E}}_2=\frac{1}{4{\mathrm{Ï€Î\mu Ï\mu }}_0}×\frac{\mathrm{q}}{{\mathrm{R}}^2}$

Since the vertical component of both the field cancels each other meaning that the net y components of the electric field at a chosen point is zero.

${\mathrm{E}}_{\mathrm{y}}=0$

However,${\mathrm{E}}_{\mathrm{x}}={\mathrm{E}}_2-2\mathrm{·¡³¦´Ç²õθ}$

Meaning that the horizontal electric field is too small. Same is the case with in the right of the negative charge at the point at same distance at this left point.

Hence, we can conclude that the horizontal line passing through the negative charge contains the point which has the least net electric field.