Point charges +4q, - q and +4q are kept on the X -axis at point x = 0 and x = 2a respectively.

Question

A 5-ohm resistor is connected to a 10 V battery. Calculate the current flowing through the resistor.

Answer 1

To solve this problem, we need to analyze the equilibrium conditions of the three-point charges placed along the X-axis at given positions.

Let's start by examining the setup:
- Charge +4q is located at position x = 0.
- Charge -q is located at an undetermined position that leads to equilibrium conditions; let's denote it as x = a.
- Another charge +4q is at position x = 2a.
The condition for the equilibrium of a charge is that the net force acting on it should be zero. In our case, due to the symmetric placement and differing signs of the charges, this scenario offers insight into their stability.
Consider each charge for equilibrium:
- Charge +4q at x = 0:
  - The forces due to the other charges will not be zero because the charges are on a straight line, and the forces from -q and +4q at x = 2a cannot balance each other to zero.
- Charge -q at x = a:
  - If placed here to maintain any form of equilibrium, it would require precise adjustment against both +4q charges on its sides — not achieving stable equilibrium as any small displacement would destabilize it.
- Charge +4q at x = 2a:
  - Similar to the +4q at x = 0, symmetry in linear positioning of similarly signed charges results in a non-zero net force unless conditions are perfectly met, which is not stable in practicality.
Because no individual charge can reach a state where it's unaffected by the nearby charges' positions unless under precise, unbalanced conditions, we conclude that they are all in a state of unstable equilibrium.

Thus, the correct answer is: all of the charges are in unstable equilibrium.

Answer 2

To solve this problem, we need to analyze the equilibrium conditions of the three-point charges placed along the X-axis at given positions.

Let's start by examining the setup:
- Charge +4q is located at position x = 0.
- Charge -q is located at an undetermined position that leads to equilibrium conditions; let's denote it as x = a.
- Another charge +4q is at position x = 2a.
The condition for the equilibrium of a charge is that the net force acting on it should be zero. In our case, due to the symmetric placement and differing signs of the charges, this scenario offers insight into their stability.
Consider each charge for equilibrium:
- Charge +4q at x = 0:
  - The forces due to the other charges will not be zero because the charges are on a straight line, and the forces from -q and +4q at x = 2a cannot balance each other to zero.
- Charge -q at x = a:
  - If placed here to maintain any form of equilibrium, it would require precise adjustment against both +4q charges on its sides — not achieving stable equilibrium as any small displacement would destabilize it.
- Charge +4q at x = 2a:
  - Similar to the +4q at x = 0, symmetry in linear positioning of similarly signed charges results in a non-zero net force unless conditions are perfectly met, which is not stable in practicality.
Because no individual charge can reach a state where it's unaffected by the nearby charges' positions unless under precise, unbalanced conditions, we conclude that they are all in a state of unstable equilibrium.

Thus, the correct answer is: all of the charges are in unstable equilibrium.

The Correct Option is C