Electrostatic Potential

Solution:

To solve this problem, let's break it down step by step.

Step 1: Understanding the electric potential due to point charges

The potential at a point due to a point charge at a distance from the charge is given by:

V =

Where:

• is the electric potential at point ,
• is Coulomb’s constant ,
• is the charge,
• is the distance from the point charge to point .

Step 2: Setup for the problem

We are given:

• A charge at the origin ,
• A charge at ,
• An equipotential circle of radius centered at where the potential .

Step 3: Equation for the total potential

The total potential at any point due to both charges is the sum of the potentials due to each charge. For a point at a distance from the charge and a distance from the charge , the total potential is:

V_total =

Since we are looking for the radius where the potential is zero, we set:

V_total = 0

Thus:

Simplifying:

Step 4: Distance relations

From the figure, the distance between the point on the equipotential circle and the two charges will be given by the Pythagorean theorem.

• The distance to the first charge at is ,
• The distance to the second charge at is .

Step 5: Solve for

Now, we use the fact that the two charges are placed symmetrically along the x-axis, and the equation for the equipotential circle, where the total potential is zero, simplifies the calculations. The radius of the circle of zero potential is given as:

R = 1.73 meters

Final Answer:

The value of is 1.73 meters.

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