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?
1/4pienot =9 *10^9 m/F
Coulomb’s Law:
F=k•q₁•q₂/r² =…
F=ma => a=F/m=…
r=at²/2
t=sqrt{2r/a)=…
no not this way its accelaration is continuously increasing so integration is required......
whats the answer
what ia answer?please help.
0.74
HOW DID U GET THAT?
To find the time at which the point charges collide, we can use the principle of conservation of mechanical energy. According to this principle, the sum of kinetic and potential energies remains constant throughout the motion.
Initially, the charges are held at fixed positions, so there is no kinetic energy. The only form of energy present is potential energy.
The potential energy between two point charges is given by the equation:
U = k * |q1 * q2| / r
Where U is the potential energy, k is the electrostatic constant (1/4πε₀ = 9 × 10^9 Nm²/C²), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
In this case, the potential energy between the charges is:
U = k * (|+q * -q|) / r
= k * (|1 μC * -1 μC|) / r
= k * (1 μC * 1 μC) / r
= k / r
Since there is no external force acting on the system, mechanical energy remains conserved throughout the motion. Initially, there is only potential energy, and when they collide, there will be only kinetic energy.
At maximum separation, when the charges are at their initial positions, all the energy is potential energy.
When the charges collide, the relative separation distance becomes zero, and all the energy is kinetic.
Therefore, we can equate the potential energy at the maximum separation to the kinetic energy at the collision:
k / r = (1/2) * (m * v^2)
Where v is the velocity of each charge at collision.
Since both charges have the same mass and velocity at collision, we can rewrite the equation as:
k / r = (1/2) * (m * v^2)
k / r = (1/2) * (2 * m * v^2)
k / r = m * v^2
Now, rearrange the equation to solve for the velocity v:
v^2 = (k / r) / m
v = sqrt((k / r) / m)
Plug in the values of k (9 × 10^9 Nm²/C²), r (1 m), and m (1 g or 0.001 kg):
v = sqrt((9 × 10^9 Nm²/C²) / (1 m * 0.001 kg))
v = sqrt(9 × 10^12 Nm²/(C² * kg))
v = 3 × 10^6 Ns/C
Now, to find the time τ it takes for the charges to collide, we need to find the distance traveled by each charge (±q) from their initial positions (±r₀) to the point of collision when they both reach the origin.
Since the charges are released from rest at t = 0, we can use the equation of motion to find the distance traveled (s) by a uniformly accelerated particle:
s = ut + (1/2) * at^2
Where u is the initial velocity (0 in this case), a is the acceleration, and t is the time taken.
The acceleration experienced by each charge due to the electrostatic repulsion can be calculated using Newton's second law:
F = ma
k * (|q|^2 / r^2) = ma
m * a = k * (|q|^2 / r^2)
a = k * (|q|^2 / (m * r^2))
Since both charges have the same mass and magnitude but opposite signs, we can rewrite the equation as:
a = k * (|q|^2 / (m * r^2))
a = k * ((1 μC)^2 / (0.001 kg * 1 m^2))
a = 9 × 10^9 Nm²/C²
Now, let's calculate the distance traveled by either charge (±q) from their initial positions to the point of collision:
s = 0 * t + (1/2) * (9 × 10^9 Nm²/C²) * t^2
s = (1/2) * (9 × 10^9 Nm²/C²) * t^2
s = 4.5 × 10^9 Nm²/C² * t^2
Since both charges have the same distance to travel, we can equate the distance traveled by either charge to the separation distance (2r₀) between their initial positions:
4.5 × 10^9 Nm²/C² * t^2 = 2 * r₀
t^2 = (2 * r₀) / (4.5 × 10^9 Nm²/C²)
t^2 = (2 * 1 m) / (4.5 × 10^9 Nm²/C²)
t^2 = 4.4 × 10^(-10) m²/N
t = sqrt(4.4 × 10^(-10) m²/N)
Finally, calculate the value of τ (time at which the charges collide) by doubling the calculated time t:
τ = 2 * sqrt(4.4 × 10^(-10) m²/N)
τ = 2 * (2.1 × 10^(-5) m/N)
τ ≈ 4.2 × 10^(-5) s
Therefore, the time at which the charges collide (τ) is approximately 4.2 × 10^(-5) seconds.