Two charges of equal magnitude of 2C and mass of 1.0 g are separated by a distance of 3 mm. Find the velocity of one particle when they are separated by a distance of 4mm.
Are both particles free to move? Was the velocity 0 when they were 3 mm apart?
If so, they move at equal velocity in opposite directions. Use conservation of energy and the change in electrostatic potential energy to compute the increase in kinetic energy.
kQ^2/0.003m - kQ^2/0.004m = 2*(1/2) M V^2
M = 0.001 kg
k is the Coulomb constant
To find the velocity of one particle when the charges are separated by a distance of 4 mm, we can use the principle of conservation of energy and the formula for electric potential energy.
The electric potential energy between two charges is given by the formula:
U = k*(q1*q2)/r
where U is the electric potential energy, k is Coulomb's constant (k = 9.0 x 10^9 N·m^2/C^2), q1 and q2 are the magnitudes of the charges, and r is the distance between the charges.
Initially, the charges are separated by a distance of 3 mm, so the electric potential energy is:
U1 = k*(2C)*(2C)/(3mm)
Now, when the charges are separated by a distance of 4 mm, the electric potential energy will be:
U2 = k*(2C)*(2C)/(4mm)
According to the principle of conservation of energy, the change in potential energy must be equal to the change in kinetic energy.
So, the change in kinetic energy is:
ΔK.E. = ΔU = U2 - U1
Now, since kinetic energy is given by the formula:
K.E. = (1/2)*m*v^2
where m is the mass and v is the velocity of the charged particle, we can rearrange the equation to solve for the velocity:
v = sqrt((2*ΔK.E.)/m)
Substituting the values, we can find the velocity of the charged particle.