To solve the problem of finding the points in the x-y plane where the electric potential due to the two fixed charges is zero, we need to start by understanding how electric potential works. The potential \( V \) at a point in space due to a point charge is given by the formula:
\( V = \frac{kQ}{r} \)
where \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point where the potential is being calculated. In this case, we have two charges: \( -2Q \) located at \( (-3a, 0) \) and \( Q \) located at \( (3a, 0) \).
Setting Up the Equation
Let’s denote a point in the x-y plane as \( P(x, y) \). The potential at point \( P \) due to the charge \( -2Q \) is:
\( V_{-2Q} = \frac{k(-2Q)}{r_{-2Q}} = \frac{-2kQ}{\sqrt{(x + 3a)^2 + y^2}} \)
And the potential at point \( P \) due to the charge \( Q \) is:
\( V_{Q} = \frac{kQ}{r_{Q}} = \frac{kQ}{\sqrt{(x - 3a)^2 + y^2}} \)
To find the points where the total potential is zero, we set the sum of these potentials to zero:
\( V_{-2Q} + V_{Q} = 0 \)
Combining the Potentials
Substituting the expressions for \( V_{-2Q} \) and \( V_{Q} \) into the equation gives us:
\( \frac{-2kQ}{\sqrt{(x + 3a)^2 + y^2}} + \frac{kQ}{\sqrt{(x - 3a)^2 + y^2}} = 0 \)
We can simplify this equation by multiplying through by the denominators to eliminate the fractions:
\( -2kQ \sqrt{(x - 3a)^2 + y^2} + kQ \sqrt{(x + 3a)^2 + y^2} = 0 \)
Dividing through by \( kQ \) (assuming \( Q \neq 0 \)) results in:
\( -2 \sqrt{(x - 3a)^2 + y^2} + \sqrt{(x + 3a)^2 + y^2} = 0 \)
Solving for the Circle Equation
Rearranging gives us:
\( \sqrt{(x + 3a)^2 + y^2} = 2 \sqrt{(x - 3a)^2 + y^2} \)
Next, squaring both sides to eliminate the square roots leads to:
\( (x + 3a)^2 + y^2 = 4((x - 3a)^2 + y^2) \)
Expanding both sides results in:
\( x^2 + 6ax + 9a^2 + y^2 = 4(x^2 - 6ax + 9a^2 + y^2) \)
Distributing the 4 gives:
\( x^2 + 6ax + 9a^2 + y^2 = 4x^2 - 24ax + 36a^2 + 4y^2 \)
Now, rearranging terms leads to:
\( 0 = 3x^2 - 30ax + 27a^2 + 3y^2 \)
Dividing through by 3 simplifies this to:
\( 0 = x^2 - 10ax + 9a^2 + y^2 \)
Identifying the Circle's Properties
This can be rearranged into the standard form of a circle:
\( (x - 5a)^2 + y^2 = 16a^2 \)
From this equation, we can see that the center of the circle is at \( (5a, 0) \) and the radius is \( 4a \).
Summary of Findings
- Center of the Circle: \( (5a, 0) \)
- Radius: \( 4a \)
Thus, all points in the x-y plane where the potential due to the two charges is zero lie on a circle centered at \( (5a, 0) \) with a radius of \( 4a \). This geometric interpretation helps visualize how the electric potentials from the two charges interact in space.
