Question

Point charge q1 = -5.00 nC is at the origin and point charge q2 = +3.00 nC is on the x-axis at x = 3.00 cm. Point P is on the y-axis at y = 4.00 cm. (a) Calculate the electric fieldsandat point P due to the charges q₁ and q₂. Express your results in terms of unit vectors (see Example 21.6). (b) Use the results of part (a) to obtain the resultant field at P, expressed in unit vector form.

Short answer

Answer:

(a) The electric field E₁ and E₂ at the point P due to the point q₁ and q₂ is ,

$E_1=\left[-2.8\times {10}^4\hat{j}\right]N/C,\left[-6.48\times {10}^3\hat{i}+8.64\times {10}^3\hat{j}\right]N/m$

(b) The resultant field at P isrole="math" localid="1658483325561" $\left[-6.48\times {10}^3\hat{i}+1.94\times {10}^4\hat{j}\right]N/m$

Step-by-step solution

Electric field vector

Given,

$$\begin{gathered} q_1=-5.0nC \\ {q}_2=3.0nC \\ x=3.0\times {10}^{-2}m \\ y=4.0\times {10}^{-2}\text{m} \end{gathered}$$

The electric field due to the charge is,

$E_1=\frac{k{q}_1}{r_1^2}$

Now, the electric field due to the charge at origin is given by:

$$\begin{gathered} E_1=\frac{k{q}_1}{r_1^2} \\ =\frac{9\times {10}^9\times 5\times {{10}^{-}}^9}{{\left(4\times {{10}^{-}}^2\right)}^2} \\ =2.8\times {10}^4 N/C \end{gathered}$$

Now the direction of the field is in

Therefore,$E_1=\left[-2.8\times {10}^4\hat{j}\right]N/C$

Now, the electric field due to the second charge is given by:

$$\begin{gathered} E_2=\frac{k{q}_2}{{r_2}^2} \\ =\frac{k{q}_2}{{\left(\sqrt{\left(x^2+y^2\right)}\right)}^2} \\ ={\frac{9\times {10}^9\times 3\times {{10}^{-}}^9}{\sqrt{{\left({\left(0.3\right)}^2+{\left(0.4\right)}^2\right)}^2}}}^2 \\ =1.08\times {10}^4 N/C \end{gathered}$$

Calculating the field vectors

Therefore, the angle and the electric field and the horizontal is:

$$\begin{gathered} {\theta }_1=arctan\left(\frac{4}{3}\right) \\ =53.{13}^0 \end{gathered}$$

Now, the electric field vector is given by:

$$\begin{gathered} E_2=E_1-\left(cos{\theta }_i+sin\theta \right) \\ =1.08\times {10}^4\left(-cos\left(53.13\right)\hat{i}+sin\left(53.13\right)\hat{j}\right) \\ =\left[-6.48\times {10}^3\hat{i}+8.64\times {10}^3\hat{j}\right]N/m \end{gathered}$$

Therefore, the resultant field at P is$\left[-6.48\times {10}^3\hat{i}+8.64\times {10}^3\hat{j}\right]N/m$

Now the net electric field is:

$$\begin{gathered} E=E_1+E_2 \\ =-2.8\times {10}^4\hat{j}-6.48\times {10}^3\hat{i}+8.64\times {10}^3\hat{j} \\ =\left[-6.48\times {10}^3\hat{i}-1.94\times {10}^4\hat{j}\right]N/m \end{gathered}$$