Two point charges +q=1 μC and −q=−1 μC with mass m=1 g are fixed at the positions ±r⃗ 0 with |r0|=1 m. The charges are released from rest at t=0. Find the time τ in seconds at which they collide.

Hint: Can you do it without integrating by using Kepler's laws?
Details and assumptions

k=14πϵ0=9×109 m/F

0.74 how?

To find the time at which the charges collide, we need to determine their trajectories and calculate when they will intersect. We can solve this problem without integrating by utilizing Kepler's laws.

Initially, let's analyze the situation. We have two point charges, +q and -q, with mass m fixed at positions ±r0. The charges are released from rest at t=0. The gravitational constant, G, is not given, so we can assume it is not applicable in this problem.

First, let's consider the forces acting on the positively charged particle, +q. It experiences an attractive electric force towards the negatively charged particle, -q. Since the charges have opposite signs, the electric force acting on +q is attractive and directed towards -q.

Now, let's analyze the motion of +q using Kepler's law of orbits. According to Kepler's first law, the trajectory of +q will be an ellipse with -q at one of its foci.

Next, we can apply Kepler's second law, which states that the area swept out by the line connecting the two charges is constant. Since +q and -q start at rest and all the force acting on +q is the attractive electric force, it will move in such a way that equal areas are swept out in equal times.

Since +q and -q have the same mass, the center of mass of the system will lie at the midpoint between them, which is the origin. Hence, we can say that the trajectories of +q and -q will be symmetrical with respect to the origin.

From this, we can infer that the time taken by +q to move from its initial position +r0 to the origin is the same as the time taken by -q to move from its initial position -r0 to the origin.

Since both particles start from opposite points, travel equal distances, and have the same acceleration, they will have the same time of flight to reach the origin or collide.

Therefore, the time at which the charges collide, denoted as τ, will be the time taken by either charge to reach the origin from their respective initial positions. Hence, we only need to find the time taken by +q to reach the origin.

To calculate this time, we can use the equation of motion for uniformly accelerated motion:

s = ut + (1/2)at^2

Since +q starts from rest, its initial velocity, u, is 0. The displacement, s, is the distance from +r0 to the origin, which is r0. And the acceleration, a, is determined by the attractive electric force between +q and -q:

F = ma

From the electric force equation:

F = k(q1q2) / r^2

Substituting the mass and charges in the equation, we have:

ma = k(q)(-q) / r0^2

Simplifying, we get:

a = -kq^2 / (m*r0^2)

Substituting the value of k and q, we can calculate the value of a.

Once we have the value of a, we can substitute it back into the equation of motion and solve for the time, t.

r0 = 1 m (given)
q = 1 μC = 1 * 10^-6 C (given)
m = 1 g = 1 * 10^-3 kg (given)
k = 9 * 10^9 m/F (given)

a = -(k * q^2) / (m * r0^2)
= -(9 * 10^9 * (1 * 10^-6)^2) / (1 * 10^-3 * 1^2)
= -9 * 10^(9-12-3)
= -9 * 10^(-6) m/s^2

Using the equation of motion:

r0 = (1/2) * a * t^2

Plugging in the values:

1 = (1/2) * (-9 * 10^-6) * t^2

Simplifying, we can solve for t:

t^2 = (2 / (-9 * 10^-6))
t^2 = -2 * (10^6 / 9)
t = sqrt(-2 * (10^6 / 9))

However, this equation gives us an imaginary value for t. This means that the charges will not collide, but rather approach each other asymptotically without ever intersecting.