Worked examples · Sheet 3

Charged particles accelerated by potential differences

Theory used: Conservation of energy for a charge in an electrostatic field Electronvolt Potential difference in a uniform field Electric potential Potential energy of two point charges

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

An electron starts from rest and is accelerated through a potential difference of $1.5 kV$ . Compute the final kinetic energy, in electronvolts and in joules, and the final speed. Neglect the weight.

\frac{1}{2} m_{e} v^{2} = e ∣Δ V ∣ ⟹ v = \frac{2 e ∣Δ V ∣}{m _{e}}

K v = 1.5 keV = 1.5 \cdot 1 0^{3} \cdot 1.60 \cdot 1 0^{- 19} J = 2.4 \cdot 1 0^{- 16} J = \frac{2 \cdot 2.4 \cdot 1 0 ^{- 16}}{9.11 \cdot 1 0 ^{- 31}} m/s \approx 2.3 \cdot 1 0^{7} m/s

The only force is the electric force, so the conservation of energy for a charge in an electrostatic field applies, and therefore the potential energy lost by the electron, $e ∣Δ V ∣$ , becomes kinetic energy. The electron is negative and moves by itself toward higher potentials, which is the direction in which its potential energy decreases. By the definition of the electronvolt a charge $e$ crossing $1.5 kV$ acquires $1.5 keV$ , and the conversion to joules is done by multiplying by $1.60 \cdot 1 0^{- 19}$ . The speed is about $7.7 %$ of the speed of light, therefore the nonrelativistic formula for the kinetic energy is still acceptable.

Exercise 2

Two flat parallel plates are $d = 2.0 cm$ apart; the first is at potential $0$ and has a small hole, the second is at potential $+ 800 V$ . A proton, of mass $m_{p} = 1.67 \cdot 1 0^{- 27} kg$ , enters through the hole with speed $v_{0} = 3.0 \cdot 1 0^{5} m/s$ perpendicular to the plates. Determine whether the proton reaches the second plate; if not, compute at what distance from the first plate it stops and with what speed it comes back out of the hole. Neglect edge effects and the weight.

The proton enters through the hole in the plate at potential

0

and the field, directed toward the hole, slows it down; it stops at

1.2 cm

from the hole and turns back

K_{0} = \frac{1}{2} m_{p} v_{0}^{2}, E = \frac{Δ V}{d}

\frac{1}{2} m_{p} v_{0}^{2} + e \cdot 0 = 0 + e E x ⟹ x = \frac{m _{p} v _{0}^{2}}{2 e E} = \frac{K _{0}}{e Δ V} d

K_{0} E x = \frac{1}{2} \cdot 1.67 \cdot 1 0^{- 27} \cdot (3.0 \cdot 1 0^{5})^{2} J \approx 7.5 \cdot 1 0^{- 17} J \approx 470 eV = \frac{800}{0.020} V/m = 4.0 \cdot 1 0^{4} V/m = \frac{7.5 \cdot 1 0 ^{- 17}}{1.60 \cdot 1 0 ^{- 19} \cdot 4.0 \cdot 1 0 ^{4}} m \approx 1.2 \cdot 1 0^{- 2} m

The field between the plates is uniform, because edge effects are negligible, and it goes from the plate at higher potential to the plate at lower potential, that is, toward the hole; a force opposite to the velocity therefore acts on the proton, which is positive. To reach the second plate the proton would have to increase its potential energy by $e \cdot 800 V = 800 eV$ , while its initial kinetic energy is about $470 eV$ ; therefore the proton does not reach the second plate.

The proton stops at the point where all its kinetic energy has become potential energy. By the potential difference in a uniform field, at distance $x$ from the hole the potential is $E x$ , and conservation of energy gives $x \approx 1.2 cm$ , that is, $59 %$ of the distance between the plates, equal to the ratio $\frac{470}{800}$ . Then the force, still directed toward the hole, pushes the proton back; at the hole the potential is again $0$ , so the kinetic energy returns to its initial value and the proton comes out with its entry speed, $3.0 \cdot 1 0^{5} m/s$ , in the opposite direction.

Exercise 3

Two small insulating spheres, of masses $m_{1} = 2.0 g$ and $m_{2} = 4.0 g$ , both carry the charge $q = 3.0 μ C$ and are held at rest at distance $r_{0} = 10 cm$ on a smooth horizontal plane. The spheres are released. Compute their speeds when the distance has become $20 cm$ and when the distance is very large.

⎩ ⎨ ⎧ m_{1} v_{1} = m_{2} v_{2} \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} = \frac{k q ^{2}}{r _{0}} - \frac{k q ^{2}}{r} ⟹ v_{1} = \frac{2 m _{2} Δ U}{m _{1} ( m _{1} + m _{2} )}, v_{2} = \frac{m _{1}}{m _{2}} v_{1}

\frac{k q ^{2}}{r _{0}} r \to + \infty : v_{1} r = 20 cm : Δ U = \frac{8.99 \cdot 1 0 ^{9} \cdot ( 3.0 \cdot 1 0 ^{- 6} ) ^{2}}{0.10} J \approx 0.809 J = \frac{2 \cdot 4.0 \cdot 1 0 ^{- 3} \cdot 0.809}{2.0 \cdot 1 0 ^{- 3} \cdot 6.0 \cdot 1 0 ^{- 3}} m/s \approx 23 m/s, v_{2} \approx 12 m/s = \frac{k q ^{2}}{r _{0}} - \frac{k q ^{2}}{2 r _{0}} = \frac{k q ^{2}}{2 r _{0}} \approx 0.405 J, v_{1} \approx 16 m/s, v_{2} \approx 8.2 m/s

The forces acting on the spheres are the weight and the normal force of the plane, which balance each other, and the electric force between the two spheres, which is an internal force; the plane is smooth, so the total momentum, initially zero, remains zero, and the spheres move in opposite directions with $m_{1} v_{1} = m_{2} v_{2}$ . The spheres are small and insulating, therefore they are treated as point charges, and the only force that does work is the electric force, which is conservative; by conservation of energy the total kinetic energy equals the decrease of the potential energy $\frac{k q ^{2}}{r}$ of the system. The two conditions, which must hold together, form the system; substituting $v_{2} = \frac{m _{1}}{m _{2}} v_{1}$ into the second equation gives $v_{1}$ .

At a distance of $20 cm$ the potential energy has halved, so half of the available energy has been released and the speeds are the final ones divided by $2$ . The lighter sphere has twice the speed, and hence twice the kinetic energy: it receives $\frac{2}{3}$ of the energy released.