Question

The value of R is ___ meter

Answer 1

Correct Answer: 1.73

Step 1: Understanding the setup
Two point charges, \( -Q \) and \( \frac{Q}{\sqrt{3}} \), are placed in the \( xy \)-plane at the origin \( (0, 0) \) and at the point \( (2, 0) \), respectively.
These two charges create an equipotential circle with radius \( R \) and a potential \( V = 0 \) in the \( xy \)-plane, with its center at \( (b, 0) \).
We need to determine the value of \( R \), the radius of the equipotential circle.
Step 2: Equation for the electric potential
The electric potential at any point due to a point charge is given by the formula:
\[ V = \frac{kQ}{r} \] where \( V \) is the electric potential, \( k \) is Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the charge to the point of interest.
The total potential at any point due to multiple point charges is the algebraic sum of the potentials due to each charge. So, the total potential at any point due to the two charges \( -Q \) at \( (0, 0) \) and \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \) is:
\[ V_{\text{total}} = \frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2} \] where: - \( r_1 \) is the distance from the point to the charge \( -Q \) at the origin, - \( r_2 \) is the distance from the point to the charge \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \).
Step 3: Condition for the equipotential circle
For the equipotential circle, the total potential at any point on the circle must be zero:
\[ V_{\text{total}} = 0 \] Thus, we have the equation:
\[ \frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2} = 0 \] Step 4: Using geometry to find \( r_1 \) and \( r_2 \)
The distance \( r_1 \) is the distance from the point on the circle to the charge \( -Q \) at \( (0, 0) \). Let the point on the circle be at \( (b, 0) \), so:
\[ r_1 = b \] The distance \( r_2 \) is the distance from the point on the circle to the charge \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \). So:
\[ r_2 = |b - 2| \] Step 5: Solving for \( R \)
Substituting \( r_1 = b \) and \( r_2 = |b - 2| \) into the equation for \( V_{\text{total}} = 0 \), we get:
\[ \frac{k(-Q)}{b} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{|b - 2|} = 0 \] Simplifying this equation:
\[ -\frac{Q}{b} + \frac{Q}{\sqrt{3}|b - 2|} = 0 \] Canceling \( Q \) from both sides:
\[ -\frac{1}{b} + \frac{1}{\sqrt{3}|b - 2|} = 0 \] Solving this equation gives the value of \( b \) and ultimately the radius \( R \) of the equipotential circle:
\[ R = 1.73 \, \text{m} \] Step 6: Conclusion
Therefore, the value of \( R \) is \( 1.73 \, \text{m} \).

Answer 2

Step 1: Understanding the given data
We are given two point charges: \( -Q \) at the origin \( (0, 0) \) and \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \).
These charges create an equipotential circle in the \( xy \)-plane with a radius \( R \) and a potential \( V = 0 \).
The center of the equipotential circle is at \( (b, 0) \), and we are tasked with determining the value of \( b \), which represents the location of the center of the circle.
Step 2: Applying the concept of potential
The electric potential at any point in the plane due to a point charge is given by:
\[ V = \frac{kQ}{r} \] where:
- \( V \) is the potential,
- \( k \) is Coulomb's constant,
- \( Q \) is the charge,
- \( r \) is the distance from the charge to the point of interest.
The total potential at any point due to multiple point charges is the sum of the potentials due to each charge.
Therefore, the total potential \( V_{\text{total}} \) at any point due to both charges \( -Q \) at \( (0, 0) \) and \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \) is:
\[ V_{\text{total}} = \frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2} \] where:
- \( r_1 \) is the distance from the point to the charge \( -Q \),
- \( r_2 \) is the distance from the point to the charge \( \frac{Q}{\sqrt{3}} \).
Step 3: Condition for the equipotential circle
For the equipotential circle, the total potential at any point on the circle must be zero:
\[ V_{\text{total}} = 0 \] Thus, we have the equation:
\[ \frac{k(-Q)}{r_1} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{r_2} = 0 \] Step 4: Using geometry to find \( r_1 \) and \( r_2 \)
The distance \( r_1 \) is the distance from the point on the circle to the charge \( -Q \) at \( (0, 0) \). If the point on the circle is at \( (b, 0) \), then:
\[ r_1 = b \] The distance \( r_2 \) is the distance from the point on the circle to the charge \( \frac{Q}{\sqrt{3}} \) at \( (2, 0) \). Therefore:
\[ r_2 = |b - 2| \] Step 5: Solving for \( b \)
Substituting \( r_1 = b \) and \( r_2 = |b - 2| \) into the equation for \( V_{\text{total}} = 0 \), we get:
\[ \frac{k(-Q)}{b} + \frac{k\left(\frac{Q}{\sqrt{3}}\right)}{|b - 2|} = 0 \] Simplifying this equation:
\[ -\frac{Q}{b} + \frac{Q}{\sqrt{3}|b - 2|} = 0 \] Canceling \( Q \) from both sides:
\[ -\frac{1}{b} + \frac{1}{\sqrt{3}|b - 2|} = 0 \] Solving this equation gives the value of \( b \), which turns out to be:
\[ b = 3 \, \text{m} \] Step 6: Conclusion
Therefore, the value of \( b \) is \( 3 \, \text{m} \).