The electric potential varies inversely with the distance. So, the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
What is electric potential?
The work done on an electric charge to shift it from infinity to a point is known as electric potential at that point. And its expression is,
[tex]V = \dfrac{kq}{r}[/tex]
here, k is the coulomb's constant.
Given data:
The magnitude of two point charges are, [tex]+2.0 \;\rm \mu C[/tex] and [tex]-6.0 \;\rm \mu C[/tex].
The location of each charge on the x-axis is -1.0 cm and +2.0 cm.
Let the third charge ( [tex]+3.0 \;\rm \mu C[/tex] ) be placed at a distance of x. Then the electric potential at origin is,
[tex]V = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}[/tex]
Since, potential at origin is zero (V = 0). Then,
[tex]0 = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times -6.0}{0.02} +\dfrac{k \times 3.0}{x}\\\\\dfrac{k \times 6.0}{0.02} = \dfrac{k \times 2.0}{0.01} +\dfrac{k \times 3.0}{x}\\\\\dfrac{6.0}{0.02} = \dfrac{2.0}{0.01} +\dfrac{3.0}{x}\\\\x = 0.03 \;\rm m =3 \;\rm cm[/tex]
Thus, we can conclude that the third charge should be placed at a distance of 3 cm from the origin on the x-axis.
Learn more about the electric potential here:
https://brainly.com/question/9383604
