Respuesta :
The net electric field is the vector sum of the components of the electric
field produced by the two charges.
The values of the magnitude and direction of the net electric field at the origin (approximate values) are;
- 131.6 N/C
- 12.6 ° above the negative x–axis
How are the net electric field magnitude and direction calculated?
The possible questions based on a similar question posted online are;
(a) The net electric field at the origin.
The electric field due to charge q₁ is given as follows;
[tex]\vec E_{1x} = \mathbf{ \dfrac{1}{4 \cdot \pi \cdot \epsilon_0} \cdot \dfrac{q_1}{\vec{r}^2_2}}[/tex]
Which gives;
[tex]\vec{E}_{1x} =\mathbf{ \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot cos\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) =-21.6 \, N/C[/tex]
[tex]\vec{E}_{1y} = \mathbf{\dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(-4 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2}} \cdot sin\left(arctan\left(\dfrac{0.8}{0.6} \right) \right) = 28.8 \, N/C[/tex]
Which gives;
[tex]\vec{E}_1 = \mathbf{21.6 \, N/C \cdot \hat x + 28.8 \, N/C \hat y}[/tex]
[tex]\vec{E}_{2x} = \dfrac{\left(9 \times 10^9 \, N \cdot m/C^2 \right) \cdot \left(6.00 \, nC \times \dfrac{10^{-9}C}{1 \, nC} \right)}{\left(1 \, m \right)^2} = 150 \, N/C[/tex]
Therefore;
[tex]\vec {E} = \left[ 21.6 \, N/C - 150 \, N/C \right] \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)[/tex]
[tex]\vec {E} = \mathbf{\left( -128.4 \, N/C \right) \left( \hat x \right) + \left(28.8 \, N/C \right) \left( \hat y \right)}[/tex]
The magnitude of the net electric field is therefore;
E = [tex]\sqrt{(-128.4^2 + 28.8^2)}[/tex] ≈ 131.6
- The magnitude of the net electric field at the origin is E ≈ 131.6 N/C
(b) The direction of the net electric field at the origin.
- [tex]The \ direction \ is \ arctan \left(\dfrac{28.8}{-128.4} \right) \approx \underline{ 12.6^{\circ}} \ above \ the \ negative \ x-axis[/tex]
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