91Ó°ÊÓ

Electric field at point charge is: $\mathit{E}\mathbf{=}\mathit{k}\frac{\mathbf{\left|}\mathbf{q}\mathbf{\right|}}{{\mathbf{r}}^{\mathbf{2}}}$

Whereis the magnitude of electric field,is the magnitude of the point charge,is the constant whose value is$8.98755×{10}^9 N.m^2/C^2$ and r is the distance from the point charge to where field is measured.

At point A $E_{1→A}=k\frac{\left|q_1\right|}{{r_1}_{→A}^2}$

$$\begin{gathered} =8.98755×{10}^9\frac{12.5×{{10}^{-}}^9}{10×{{10}^{-}}^2} \\ =1.1234×{10}^4 N/C \end{gathered}$$

Now,

$$\begin{gathered} {E_2}_{→A}=8.98755×{10}^9\frac{6.5×{{10}^{-}}^9}{15×{{10}^{-}}^2} \\ =2.496×{10}^3 N/C \end{gathered}$$

Therefore, the total electric field is:

$$\begin{gathered} {E_t}_{otal}={E_1}_A+{E_2}_A \\ =8.73×{10}^3 N/C \end{gathered}$$

Therefore, the direction is to the right.

Similarly, as we have done above just plugging in the value we get:

$$\begin{gathered} {E_1}_{→B}=8.98755×{10}^9\frac{12.5×{{10}^{-}}^9}{35×{{10}^{-}}^2} \\ =917.1N/C \end{gathered}$$

The electric field produced by a negative charge will point toward it.

$$\begin{gathered} {E_{2→}}_B=8.98755×{10}^9\frac{6.5×{{10}^{-}}^9}{10×{{10}^{-}}^2} \\ =5.617×{10}^3 N/C \end{gathered}$$

Therefore, the total electric field is:

$$\begin{gathered} {E_t}_{otal}=917.1+5.617×{10}^3 \\ =6.534×{10}^3 N/C \end{gathered}$$

Therefore, the electric field is to the right.

We know that

$$\begin{gathered} {q_p}_{roton}=1.60217×{{10}^{-}}^{19}C \\ {E}_A=8.737×{10}^{3}N/C \end{gathered}$$

Now,$E=\frac{F}{{q_p}_{roton}}$

Putting the values, we get:

$$\begin{gathered} F=E{q_p}_{roton} \\ =(8.737×{10}^3)(1.602×{{10}^{-}}^{1}9) \\ =1.4×{{10}^{-}}^{1}5N \end{gathered}$$

The proton is positive therefore the direction will be in the same direction of the electric field.