Respuesta :
Answer:
The electric field will be zero at x = ± ∞.
Explanation:
Suppose, A -2.0 nC charge and a +2.0 nC charge are located on the x-axis at x = -1.0 cm and x = +1.0 cm respectively.
We know that,
The electric field is
[tex]E=\dfrac{kq}{r^2}[/tex]
The electric field vector due to charge one
[tex]\vec{E_{1}}=\dfrac{kq_{1}}{r_{1}^2}(\hat{x})[/tex]
The electric field vector due to charge second
[tex]\vec{E_{2}}=\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})[/tex]
We need to calculate the electric field
Using formula of net electric field
[tex]\vec{E}=\vec{E_{1}}+\vec{E_{2}}[/tex]
[tex]\vec{E_{1}}+\vec{E_{2}}=0[/tex]
Put the value into the formula
[tex]\dfrac{kq_{1}}{r_{1}^2}(\hat{x})+\dfrac{kq_{2}}{r_{2}^2}(-\hat{x})=0[/tex]
[tex]\dfrac{kq_{1}}{r_{1}^2}(\hat{x})=\dfrac{kq_{2}}{r_{2}^2}(\hat{x})[/tex]
[tex](\dfrac{r_{2}}{r_{1}})^2=\dfrac{q_{2}}{q_{1}}[/tex]
[tex]\dfrac{r_{2}}{r_{1}}=\sqrt{\dfrac{q_{2}}{q_{1}}}[/tex]
Put the value into the formula
[tex]\dfrac{2.0+x}{x}=\pm\sqrt{\dfrac{2.0}{2.0}}[/tex]
[tex]2.0+x=x[/tex]
If x = ∞, then the equation is be satisfied.
Hence, The electric field will be zero at x = ± ∞.
At positions to the left on the x-axis is the electric field zero.
What position on the x-axis is the electric field zero?
We know that the electric fields add together in between the charges and we also know that the magnitude of the negative charge is greater than the positive charge.
So the electric field will not be zero to the right of the negative charge but on the left of the charge so we can conclude that at positions to the right on the x-axis is the electric field zero.
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