Two fixed charges -2Q and Q are located at the point with coordinates (-3a,0) and (+3a,0) respectively in the x- y plane .
Show that all points in the x-y plane where the electric potential due to the two charge is zero , lie on a circle. If its radius is r = xa and the location of its centre is (ya,0) find x and y .

Explanation

To solve the problem step by step, we need to find the points in the x-y plane where the electric potential due to the two charges is zero. The two charges are located at (-3a, 0) and (3a, 0) with magnitudes -2Q and Q respectively.

### Step 1: Define the Electric Potential
The electric potential \( V \) at a point \( (x, y) \) due to a point charge \( Q \) is given by:
\[ V = k \frac{Q}{r} \]
where \( r \) is the distance from the charge to the point where the potential is being calculated, and \( k \) is Coulomb's constant.

### Step 2: Calculate Distances
Let \( r_1 \) be the distance from the charge -2Q at (-3a, 0) to the point (x, y):
\[ r_1 = \sqrt{(x + 3a)^2 + y^2} \]
Let \( r_2 \) be the distance from the charge Q at (3a, 0) to the point (x, y):
\[ r_2 = \sqrt{(x - 3a)^2 + y^2} \]

### Step 3: Write the Expression for Total Potential
The total electric potential \( V \) at the point (x, y) due to both charges is:
\[ V = k \frac{Q}{r_2} - k \frac{2Q}{r_1} \]
Setting \( V = 0 \) gives:
\[ k \frac{Q}{r_2} - k \frac{2Q}{r_1} = 0 \]
This simplifies to:
\[ \frac{Q}{r_2} = \frac{2Q}{r_1} \]
Cancelling \( Q \) (assuming \( Q \neq 0 \)):
\[ \frac{1}{r_2} = \frac{2}{r_1} \]

### Step 4: Cross-Multiply and Simplify
Cross-multiplying gives:
\[ r_1 = 2r_2 \]
Substituting the expressions for \( r_1 \) and \( r_2 \):
\[ \sqrt{(x + 3a)^2 + y^2} = 2 \sqrt{(x - 3a)^2 + y^2} \]

### Step 5: Square Both Sides
Squaring both sides eliminates the square roots:
\[ (x + 3a)^2 + y^2 = 4((x - 3a)^2 + y^2) \]
Expanding both sides:
\[ x^2 + 6ax + 9a^2 + y^2 = 4(x^2 - 6ax + 9a^2 + y^2) \]
This simplifies to:
\[ x^2 + 6ax + 9a^2 + y^2 = 4x^2 - 24ax + 36a^2 + 4y^2 \]

### Step 6: Rearranging Terms
Rearranging gives:
\[ 0 = 3x^2 - 30ax + 3y^2 + 27a^2 \]
Dividing through by 3:
\[ 0 = x^2 - 10ax + y^2 + 9a^2 \]

### Step 7: Completing the Square
Completing the square for the x terms:
\[ (x^2 - 10ax) + y^2 + 9a^2 = 0 \]
\[ (x - 5a)^2 + y^2 = 16a^2 \]

### Step 8: Identify Circle Parameters
This is the equation of a circle centered at \( (5a, 0) \) with radius \( 4a \).

### Step 9: Relate to Given Variables
According to the problem, the center is \( (ya, 0) \) and the radius is \( xa \). Thus:
- \( y = 5 \)
- \( x = 4 \)

### Final Answer
The values of \( x \) and \( y \) are:
\[ x = 4, \quad y = 5 \]