Question

Consider two point particles that have charge +e, are at rest, and are separated by 2.7 x 10-15 m.

1) How much work was required to bring them together from a very large separation distance? keV

2) If they are released, how much kinetic energy will each have when they are separated by twice their separation at release? keV

3) The mass of each particle is 1.00 u (1.00 amu). What speed will each have when they are very far from each other? m/s

Answer 1

1) The work required to bring the charges together can be calculated using the formula for potential energy. 2) When the charges are released, each will have half of the initial potential energy when they are separated by twice their separation at release. 3) When the charges are very far from each other, they will each have the same kinetic energy as the initial potential energy.

Step-by-step explanation:

1) How much work was required to bring them together from a very large separation distance? keV

To calculate the work required to bring the two point particles together, we can use the formula:

Work = Potential Energy

The potential energy between two point charges can be calculated using the formula:

Potential Energy = (k * q1 * q2) / r

Where k is the electrostatic constant (9 x 10^9 Nm^2/C^2), q1 and q2 are the charges of the particles (+e in this case), and r is the separation distance.

Plugging in the values, we get:

Potential Energy = (9 x 10^9 * e * e) / (2.7 x 10^-15)

Converting the potential energy to keV:

Work = Potential Energy * (1 keV / 1.62 x 10^-16 J)

2) If they are released, how much kinetic energy will each have when they are separated by twice their separation at release? keV

When the particles are released, their potential energy is converted into kinetic energy. Since the kinetic energy is conserved, the total initial potential energy will be equal to the total final kinetic energy for both particles. Therefore, each particle will have half of the initial potential energy when they are separated by twice their separation at release.

3) The mass of each particle is 1.00 u (1.00 amu). What speed will each have when they are very far from each other? m/s

When the particles are very far from each other, their potential energy is zero. Using the conservation of mechanical energy, the sum of their kinetic energies is equal to the initial potential energy. Therefore, each particle will have the same kinetic energy as the initial potential energy. Using the formula:

Kinetic Energy = (1/2) * mass * velocity^2

We can solve for the velocity:

velocity = sqrt((2 * Kinetic Energy) / mass)

Answer 2

Step-by-step explanation:

q1 = q2 = e = 1.6 x 10^-19 C

d = 2.7 x 10^-15 m

1. Work done is equal to the potential energy stored between the two charges.

The formula for the potential energy is given by

`U=(Kq_(1)q_(2))/(d)`

By substituting the values

`U=(9* 10^(9)* 1.6* 10^(-19)* 1.6* 10^(-19))/(2.7* 10^(-15))`

U = 8.53 x 10^-14 J

So, the work done is 8.53 x 10^-14 J.

(2) d = 2 x 2.7 x 10^-15 m = 5.4 x 10^-15 m

So, the potential energy is

`U=(Kq_(1)q_(2))/(d)`

By substituting the values

`U=(9* 10^(9)* 1.6* 10^(-19)* 1.6* 10^(-19))/(5.4* 10^(-15))`

U = 4.27 x 10^-14 J

(3) mass, m = 1 u = 1.67 x 10^-27 kg

So, the kinetic energy is equal to the potential energy.

Let v be the velocity.

`2 * (1)/(2)mv^(2)=8.53* 10^-14`

`2 * (1)/(2)* 1.67* 10^(-27)* v^(2)=8.53* 10^(-14)`

v = 7.14 x 10^6 m/s