Question

Two-point charges are separated by 25.0 cm (Fig. E21.43). Find the net electric field these charges produce at (a) point A and (b) point B. (c) What would be the magnitude and direction of the electric force this combination of charges would produce on a proton at A?

Short answer

Answer:

(a) The net electric field these charges will produced at point A isto the right$8.73\times {10}^3$

(b) The electric charge the charges will produce at point B is$6.534\times {10}^3 N/C$

(c) The magnitude of the charge is$1.4\times {{10}^{-}}^{15}N$, to the right.

Step-by-step solution

Net electric field produced at point A

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$8.98755\times {10}^9 N.m^2/C^2$ and r is the distance from the point charge to where field is measured.

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

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

Now,

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

Therefore, the total electric field is:

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

Therefore, the direction is to the right.

Electric field at point B

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

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

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

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

Therefore, the total electric field is:

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

Therefore, the electric field is to the right.

Magnitude of the proton

We know that

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

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

Putting the values, we get:

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

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