A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at a speed of 6.20 105 m/s enters an electric field, traveling from point circle a, at a potential of 1.50 103 V, to point circle b, at 4.00 103 V. What is its speed at point circle b?

To calculate the speed of the helium nucleus at point B, we can use the conservation of energy. The change in electric potential energy will be converted into kinetic energy.

The electric potential energy at point A is given by:
PEa = q * Va
where q is the charge of the helium nucleus and Va is the potential at point A.

The electric potential energy at point B is given by:
PEb = q * Vb
where Vb is the potential at point B.

The change in potential energy (ΔPE) is then:
ΔPE = PEb - PEa = q * Vb - q * Va = q * (Vb - Va)

We can equate the change in potential energy to the change in kinetic energy:
ΔPE = ΔKE
q * (Vb - Va) = 0.5 * m * (vf^2 - vi^2)

Rearranging the equation, we get:
vf^2 = vi^2 + (2 * q * (Vb - Va) / m)

Substituting the values:
vf^2 = (6.20 * 10^5 m/s)^2 + (2 * (2e) * (4.00 * 10^3 V - 1.50 * 10^3 V) / 6.63 * 10^-27 kg)

Calculating:
vf^2 = 3.84 * 10^11 m^2/s^2 + (2 * 2 * 1.50 * 10^3 V / 6.63 * 10^-27 kg - 2 * 2 * 4.00 * 10^3 V / 6.63 * 10^-27 kg)
vf^2 = 3.84 * 10^11 m^2/s^2 + (3.03 * 10^15 m^2/s^2 - 8.04 * 10^15 m^2/s^2)
vf^2 = 3.84 * 10^11 m^2/s^2 - 5.01 * 10^15 m^2/s^2
vf^2 = - 4.61 * 10^15 m^2/s^2

Since we cannot have a negative speed, it means that there is not enough energy for the helium nucleus to reach point B.
Thus, the helium nucleus will slow down and come to a stop before reaching point B.

To find the speed of the helium nucleus at point b, we can make use of the conservation of energy principle. The total energy of the nucleus can be expressed as the sum of its kinetic energy (KE) and its potential energy (PE).

The kinetic energy of the nucleus is given by the equation KE = (1/2)mv^2, where m is the mass of the nucleus and v is its speed.

The potential energy of the nucleus in an electric field is given by the equation PE = qV, where q is the charge on the nucleus and V is the potential difference across the field.

At point A, the potential is 1.50 x 10^3 V, and at point B, the potential is 4.00 x 10^3 V. The potential difference across the field is then ΔV = VB - VA = 4.00 x 10^3 V - 1.50 x 10^3 V = 2.50 x 10^3 V.

Since the charge on the helium nucleus is 2e, and the electron charge e is 1.60 x 10^-19 C, we can calculate the potential energy difference as PE = (2e)(ΔV) = (2)(1.60 x 10^-19 C)(2.50 x 10^3 V).

Now, equate the total energy at points A and B:

KE + PE at A = KE + PE at B

(1/2)mvA^2 + 2eΔV = (1/2)mvB^2

We need to find the speed vB at point B.

To proceed, note that the mass m of the helium nucleus is given as 6.63 x 10^-27 kg, and the speed vA at point A is given as 6.20 x 10^5 m/s.

Rearranging the equation gives:

(1/2)mvA^2 + 2eΔV = (1/2)mvB^2

Substituting the known values, we have:

(1/2)(6.63 x 10^-27 kg)(6.20 x 10^5 m/s)^2 + 2(2)(1.60 x 10^-19 C)(2.50 x 10^3 V) = (1/2)(6.63 x 10^-27 kg)(vB^2)

Simplifying and solving for vB gives:

(1/2)(6.63 x 10^-27 kg)(6.20 x 10^5 m/s)^2 + 2(2)(1.60 x 10^-19 C)(2.50 x 10^3 V) = (1/2)(6.63 x 10^-27 kg)(vB^2)

vB^2 = [(1/2)(6.63 x 10^-27 kg)(6.20 x 10^5 m/s)^2 + 2(2)(1.60 x 10^-19 C)(2.50 x 10^3 V)] / (1/2)(6.63 x 10^-27 kg)

Finally, we can solve for vB by taking the square root of the expression on the right-hand side of the equation.

Kinetic energy increase = (potential increase) * charge

Use the final K.E and helium nucleus mass to get the velcoity

A helium nucleus ( charge2e mass