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 $R$ is _______ meter

Answer 1

Correct Answer: 1.73

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 \( P(x, y) \) due to a point charge \( Q \) at a distance \( r \) from the charge is given by:

V = \(\frac{kQ}{r}\)

Where:

• \( V \) is the electric potential at point \( P \),
• \( k \) is Coulomb’s constant \( (k = 9 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \),
• \( Q \) is the charge,
• \( r \) is the distance from the point charge to point \( P \).

Step 2: Setup for the problem

We are given:

• A charge \( -Q \) at the origin \( (0, 0) \),
• A charge \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \),
• An equipotential circle of radius \( R \) centered at \( (b, 0) \) where the potential \( V = 0 \).

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 \( r_1 \) from the charge \( -Q \) and a distance \( r_2 \) from the charge \( \frac{Q}{\sqrt{3}} \), the total potential is:

V_total = \(\frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2}\)

Since we are looking for the radius \( R \) where the potential is zero, we set:

V_total = 0

Thus:

\(\frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2} = 0\)

Simplifying:

\(\frac{-Q}{r_1} + \frac{Q}{\sqrt{3}r_2} = 0\)

Step 4: Distance relations

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

• The distance to the first charge \( -Q \) at \( (0, 0) \) is \( r_1 = \sqrt{x^2 + y^2} \),
• The distance to the second charge \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \) is \( r_2 = \sqrt{(x - 2)^2 + y^2} \).

Step 5: Solve for \( R \)

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 \( R \) of the circle of zero potential is given as:

R = 1.73 meters

Final Answer:

The value of \( R \) is 1.73 meters.