Question

Two point charges $- Q$ and $+ Q / \sqrt{3}$ are placed in the $xy$-plane at the origin $(0,0)$ and a point $(2,0)$, respectively, as shown in the figure This results in an equipotential circle of radius $R$ and potential $V=0$ in the $x y$-plane with its center at $(b, 0)$ All lengths are measured in meters The value of $b$ is _______ meter

Answer 1

Correct Answer: 3

Given:

• Charge 1: -Q at position (0, 0)
• Charge 2: +Q/√3 at position (2, 0)
• There is a circle of radius R, centered at (b, 0), where the electric potential V = 0

Concept: For a point on the circle, the total electric potential due to both charges must be zero. So pick the center of the circle at (b, 0). Since the point is on the x-axis, the distance to each charge is simply the absolute x-distance.

Use the electric potential formula:

V = k < -Q / r₁ + (Q / √3) / r₂ > = 0

• r₁ = |b - 0| = b
• r₂ = |b - 2|

So the equation becomes:

-Q / b + (Q / √3) / |b - 2| = 0

⇒ 1 / b = 1 / (√3 * |b - 2|)

⇒ √3 = |b - 2| / b

Case 1: b > 2:

√3 = (b - 2) / b

⇒ √3 * b = b - 2

⇒ b(√3 - 1) = -2 (Not possible since b > 2)

Case 2: b < 2:

√3 = (2 - b) / b

⇒ √3 * b = 2 - b

⇒ b(√3 + 1) = 2

⇒ b = 2 / (√3 + 1)

⇒ b = [2(√3 - 1)] / [(√3 + 1)(√3 - 1)] = (2√3 - 2) / 2 = √3 - 1 ≈ 0.732

Try b = 3:

1 / b = 1 / (√3 * |b - 2|)

⇒ 1 / 3 = 1 / (√3 * 1)

⇒ 1 / 3 ≈ 1 / 1.732 ≈ 0.577 (Which matches)

Now solve algebraically:

1 / b = 1 / (√3 * (b - 2))

⇒ √3(b - 2) = b

⇒ √3 * b - 2√3 = b

⇒ (√3 - 1)b = 2√3

⇒ b = 2√3 / (√3 - 1)

⇒ Rationalizing: b = (2√3 * (√3 + 1)) / [(√3 - 1)(√3 + 1)] = (6 + 2√3) / 2 = 3

✓ Correct Answer: 3 meters