91影视

The above graph shows all the possible outcomes.

Case1:

Vertical component: There should be a vertical component and have a value in upward direction.

Horizontal component: May exist in any one and the component is larger than other

Comparing with the previous prediction we get the case it not right as the net field has only vertical component downwards.

Case2:

Vertical component: There should be a vertical component and have a value in downward direction.

Horizontal component: May exist in any one and the component is larger than other

Comparing with the previous prediction we get the case it may be correct as the net field has only vertical component downwards.

Case3:

Vertical component: may exist in case of one vertical component is larger than other

Horizontal component: Must exist with value in left direction

Comparing with the previous prediction we get the case is not right as the net field has only vertical component downwards.

Case4:

Vertical component: may exist in case of one vertical component is larger than other

Horizontal component: Must exist with value in right direction

Comparing with the previous prediction we get the case is not right as the net field has only vertical component downwards.

Therefore, we conclude that$q_1$and$q_2$should be both negative.

From the above case 2 we have seen that it should be the right one and horizontal components must be equal in magnitude and opposite in direction

The angle $$\begin{gathered} E=E_1\sin \left(67.38\right)+E_2\left(22.62\right) \\ =1.1脳{10}^7\sin \left(67.38\right)+1.25脳{10}^7\sin \left(22.62\right) \\ =1.47脳{10}^7\mathrm{N}/\mathrm{C} \end{gathered}$$between $F_1$and horizontal is:

$$\begin{gathered} {胃}_1=\arccos \left(\frac{5}{13}\right) \\ ={67.38}^{鈭�} \end{gathered}$$

${胃}_2$and $E_2$is:

$$\begin{gathered} {胃}_2=\arccos \left(\frac{12}{13}\right) \\ =22.62掳 \end{gathered}$$

Horizontal component is: $E_1\cos \left(67.38\right)=E_2\cos \left(22.62\right)$

The Force $F_1$is given by:

localid="1668249114947" $$\begin{gathered} E_1=\frac{k{q}_1}{{\left(r_{1p}\right)}^2} \\ =\frac{9脳{10}^9脳3.00脳{10}^{鈭�6}}{(0.5{)}^2} \\ =1.1脳{10}^7\mathrm{N}/\mathrm{C} \end{gathered}$$

Substituting values, we get:

$$\begin{gathered} E_2=\frac{1.1脳{10}^7脳\cos \left(67.38\right)}{\cos \left(22.62\right)} \\ =4.6脳{10}^6\mathrm{N}/\mathrm{C} \end{gathered}$$

Now,

$$\begin{gathered} E=E_1\sin \left(67.38\right)+E_2\left(22.62\right) \\ =1.1脳{10}^7\sin \left(67.38\right)+1.25脳{10}^7\sin \left(22.62\right) \\ =1.47脳{10}^7\mathrm{N}/\mathrm{C} \end{gathered}$$

Therefore, the magnitude of the electric field is localid="1668249206264" $1.47脳{10}^7\mathrm{N}/\mathrm{C}$