Two point charges of +Q each have been placed at the positions (–a/2, 0, 0) and (a/2, 0, 0). The locus of the points in YZ plane where –Q charge can be placed such the that total electrostatic potential energy of the system can become equal to zero, is represented by which of the following equations ?
Answers
Given:
Two point charges of +Q each have been placed at the positions (–a/2, 0, 0) and (a/2, 0, 0).
To find:
The locus of the points in YZ plane where –Q charge can be placed such the that total electrostatic potential energy of the system can become equal to zero, is represented by which of the following equations ?
Solution:
Let "h" be the distance between a point and the x plane.
Let "x" be the distance from the origin where the plane lies.
The potential at a point due to given charge Q1 and Q2 is given by,
V = 1/4π∈₀ Q/r1 + 1/4π∈₀ (-Q)/r2
⇒ V = k Q/r1 + k (-Q)/r2
If the net potential at this point is zero, then.
0 = k Q/r1 + k (-Q)/r2
where, r1 = √[h² + (x+a/2)²]
r2 = √[h² + (x-a/2)²]
∴ 0 = k Q/√[h² + (x+a/2)²] + k (-Q)/√[h² + (x-a/2)²]
further solving the above equation, we get,
k Q/√[h² + (x+a/2)²] = k (Q)/√[h² + (x-a/2)²]
⇒ √[h² + (x-a/2)²] = √[h² + (x+a/2)²]
taking square on both the sides, we get,
⇒ h² + (x-a/2)² = h² + (x+a/2)²
⇒ (x-a/2)² = (x+a/2)²
x² + a²/2 - ax = x² + a²/2 + ax
⇒ 2ax = 0
∴ x = 0
The equation of required plane is y-z plane, x = 0.