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
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
Correct Answer: 1.73
Solution and Explanation
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 Coulombβs 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:
Vtotal = \(\frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2}\)
Since we are looking for the radius \( R \) where the potential is zero, we set:
Vtotal = 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.
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Concepts Used:
Electrostatic Potential
The electrostatic potential is also known as the electric field potential, electric potential, or potential drop is defined as βThe amount of work that is done in order to move a unit charge from a reference point to a specific point inside the field without producing an acceleration.β
SI Unit of Electrostatic Potential:
SI unit of electrostatic potential - volt
Other units - statvolt
Symbol of electrostatic potential - V or Ο
Dimensional formula - ML2T3I-1
Electric Potential Formula:
The electric potential energy of the system is given by the following formula:
U = 1/(4ΟΡº) Γ [q1q2/d]
Where q1 and q2 are the two charges that are separated by the distance d.