The charge to the left in the figure above has a magnitude of 5.50 nC, and charge to the right has a magnitude of 1.375 nC. The distance between them d is 6.50 m. Where can you place an electron along x other than at infinity such that it is in static equilibrium? Take the origin of coordinates at the left charge.x =:

To solve this problem, we need to set the forces acting on the electron due to each charge equal to each other. The force between two charges is given by Coulomb's law:

F = k * |q1*q2| / r^2

where:

• F is the force,
• k is Coulomb's constant (8.99 * 10^9 N m^2/C^2),
• q1 and q2 are the charges,
• r is the distance between the charges.

The electron can be in equilibrium in two places: one between the two charges and one to the right of the right charge.

1. For the position between the two charges (0 < x < d):

The forces due to the two charges are equal, so:

k * |q1q3| / x^2 = k * |q2q3| / (d-x)^2

where:

• q1 is the charge of the left charge (5.50 nC = 5.50 * 10^-9 C),
• q2 is the charge of the right charge (1.375 nC = 1.375 * 10^-9 C),
• q3 is the charge of the electron (-1.6 * 10^-19 C),
• d is the distance between the two charges (6.50 m),
• x is the distance from the left charge to the electron.

Solving this equation for x gives the position of the electron.

2. For the position to the right of the right charge (x > d):

The forces due to the two charges are equal, so:

k * |q1q3| / (d+x)^2 = k * |q2q3| / x^2

Solving this equation for x gives the position of the electron.