Answer:
The point at which the electrical potential is zero is x = +0.33 m.
Explanation:
By definition the electrical potential is:
[tex] V_{E} = \frac{K*q}{r} [/tex]
Where:
K: is Coulomb's constant = 9x10⁹ N*m²/C²
q: is the charge
r: is the distance
The point at which the electrical potential is zero can be calculated as follows:
[tex] V_{1} + V_{2} = 0 [/tex]
[tex] K(\frac{q_{1}}{r_{1}} + \frac{q_{2}}{r_{2}}) = 0 [/tex] (1)
q₁ is the first charge = +3 mC
r₁ is the distance from the point to the first charge
q₂ is the first charge = -6 mC
r₂ is the distance from the point to the second charge
By replacing r₁ = 1 - r₂ into equation (1) we have:
[tex]K(\frac{q_{1}}{1 - r_{2}} + \frac{q_{2}}{r_{2}}) = 0[/tex] (2)
By solving equation (2) for r₂:
[tex]r_{2} = \frac{q_{1}}{q_{1} - q_{2}} = \frac{3 mC}{3 mC - (-6 mC)} = +0.33 m[/tex]
Therefore, the point at which the electrical potential is zero is x = +0.33 m.
I hope it helps you!
Since the two charges are placed on the x-axis. Then, the electric potential is zero at the distance of 0.66 m from the second point charge on the x-axis.
The electric potential at any point is the amount of work needed to move a unit positive charge from infinity to that point.
Given data-
The magnitude of a point charge is, Q = + 3 C.
The magnitude of the second point charge is, Q' = -6 C.
The distance between the two charges is, x = 1.0 m.
The expression for the net electric potential due to both the charges is,
[tex]V = \dfrac{k \times Q}{x} +\dfrac{k \times Q'}{x-y}[/tex]
here,
k is the Coulomb's constant.
y is the distance from the second point charge on the x-axis, where the net potential is zero.
Now, for zero net electric potential due to both the charges we have,
[tex]0 = \dfrac{k \times Q}{x-y} +\dfrac{k \times Q'}{y}\\\\\\-\dfrac{k \times Q'}{y}=\dfrac{k \times Q}{x-y}[/tex]
Solving as,
[tex]-\dfrac{(-6)}{y}=\dfrac{3}{1-y}\\\\6-6y=3y\\\\y =0.66\;\rm m[/tex]
Thus, we can conclude that the electric potential is zero at the distance of 0.66 m from the second point charge on the x-axis.
Learn more about the electric potential here:
https://brainly.com/question/13296426
