Worked examples · Sheet 3

Motion of charges in a uniform electric field

Theory used: Acceleration of a charge in a uniform field Deflection of a charge in a uniform field Force on a charge placed in the field linee e traiettorie

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A proton, of mass $m_{p} = 1.67 \cdot 1 0^{- 27} kg$ and charge $e$ , starts from rest in a uniform electric field of magnitude $E = 4.0 \cdot 1 0^{4} N/C$ . Compute the acceleration, the speed after a distance $Δ x = 2.0 cm$ and the time taken to cover it, and compare the electric force with the weight.

a = \frac{e E}{m _{p}}, v = 2 a Δ x, t = \frac{v}{a}

a v t \frac{e E}{m _{p} g} = \frac{1.60 \cdot 1 0 ^{- 19} \cdot 4.0 \cdot 1 0 ^{4}}{1.67 \cdot 1 0 ^{- 27}} m/ s^{2} \approx 3.8 \cdot 1 0^{12} m/ s^{2} = 2 \cdot 3.83 \cdot 1 0^{12} \cdot 0.020 m/s \approx 3.9 \cdot 1 0^{5} m/s = \frac{3.92 \cdot 1 0 ^{5}}{3.83 \cdot 1 0 ^{12}} s \approx 1.0 \cdot 1 0^{- 7} s = \frac{6.4 \cdot 1 0 ^{- 15} N}{1.64 \cdot 1 0 ^{- 26} N} \approx 3.9 \cdot 1 0^{11}

The field is uniform, so by the theorem on the acceleration of a charge in a uniform field the proton has constant acceleration $\frac{e E}{m _{p}}$ , in the direction of the field because the charge is positive. The initial velocity is zero, therefore the motion is uniformly accelerated rectilinear motion along a field line, and the speed after the distance $Δ x$ is obtained from $v^{2} = 2 a Δ x$ ; the time follows from $v = a t$ . The ratio between the electric force and the weight is of the order of $1 0^{11}$ , so neglecting the weight is fully justified.

Exercise 2

An electron enters with velocity $v_{0} = 5.0 \cdot 1 0^{6} m/s$ a region in which there is a uniform field $E = 1.0 \cdot 1 0^{3} N/C$ , parallel to the velocity and with the same orientation. Compute the distance that the electron travels before stopping and after how much time it passes again through the entry point.

a = \frac{e E}{m _{e}}, 0 = v_{0}^{2} - 2 a d ⟹ d = \frac{v _{0}^{2}}{2 a} = \frac{m _{e} v _{0}^{2}}{2 e E}, t_{r} = \frac{2 v _{0}}{a}

a d t_{r} = \frac{1.60 \cdot 1 0 ^{- 19} \cdot 1.0 \cdot 1 0 ^{3}}{9.11 \cdot 1 0 ^{- 31}} m/ s^{2} \approx 1.76 \cdot 1 0^{14} m/ s^{2} = \frac{( 5.0 \cdot 1 0 ^{6} ) ^{2}}{2 \cdot 1.76 \cdot 1 0 ^{14}} m \approx 7.1 \cdot 1 0^{- 2} m = \frac{2 \cdot 5.0 \cdot 1 0 ^{6}}{1.76 \cdot 1 0 ^{14}} s \approx 5.7 \cdot 1 0^{- 8} s

The electron has negative charge, so the force $- e E$ is opposite to the field and therefore opposite to the velocity: the electron slows down with constant deceleration of magnitude $\frac{e E}{m _{e}}$ . The motion is uniformly accelerated rectilinear motion, and from the relation $v^{2} = v_{0}^{2} - 2 a d$ with zero final speed we obtain the stopping distance, about $7.1 cm$ .

After stopping the force keeps acting in the same direction, so the electron turns back. The velocity is $v (t) = v_{0} - a t$ and the position $v_{0} t - \frac{1}{2} a t^{2}$ returns to zero for $t_{r} = \frac{2 v _{0}}{a}$ , the instant at which the velocity equals $- v_{0}$ : the electron passes again through the entry point after $5.7 \cdot 1 0^{- 8} s$ , with speed $v_{0}$ and opposite orientation.

Exercise 3

An electron with horizontal velocity $v_{0} = 3.0 \cdot 1 0^{7} m/s$ enters halfway between two horizontal parallel plates of length $ℓ = 5.0 cm$ and $1.0 cm$ apart, between which there is a uniform vertical field of magnitude $E = 1.2 \cdot 1 0^{4} N/C$ , directed downward. A vertical screen is at $D = 20 cm$ from the exit of the plates. Verify that the electron leaves the plates without hitting them and compute the vertical displacement $Y$ of the point at which it hits the screen.

Diagram not to scale: between the plates the electron follows an arc of parabola, then it continues in a straight line to the screen

a = \frac{e E}{m _{e}}, y_{u} = \frac{a ℓ ^{2}}{2 v _{0}^{2}}, tan α = \frac{a ℓ}{v _{0}^{2}}, Y = y_{u} + D tan α

a y_{u} tan α Y = \frac{1.60 \cdot 1 0 ^{- 19} \cdot 1.2 \cdot 1 0 ^{4}}{9.11 \cdot 1 0 ^{- 31}} m/ s^{2} \approx 2.11 \cdot 1 0^{15} m/ s^{2} = \frac{2.11 \cdot 1 0 ^{15} \cdot 0.05 0 ^{2}}{2 \cdot ( 3.0 \cdot 1 0 ^{7} ) ^{2}} m \approx 2.9 \cdot 1 0^{- 3} m = \frac{2.11 \cdot 1 0 ^{15} \cdot 0.050}{( 3.0 \cdot 1 0 ^{7} ) ^{2}} \approx 0.117 = (0.0029 + 0.20 \cdot 0.117) m \approx 2.6 \cdot 1 0^{- 2} m

The field is directed downward and the electron is negative, so the force is directed upward, toward the positive plate, and the constant acceleration equals $\frac{e E}{m _{e}}$ . The initial velocity is perpendicular to the field, therefore the theorem on the deflection of a charge in a uniform field applies: between the plates the trajectory is an arc of parabola and at the exit the vertical displacement is $y_{u} \approx 2.9 mm$ . The electron entered halfway, $5.0 mm$ from each plate, and since $2.9 mm < 5.0 mm$ it leaves without hitting the upper plate.

Outside the plates the field is zero, so the electron continues in a straight line with the exit velocity, inclined at $α \approx 6. 7^{\circ}$ to the horizontal. Over the distance $D$ to the screen it rises by a further $D tan α$ , and the total displacement on the screen is $Y \approx 2.6 cm$ . The contribution of the straight stretch prevails, because $D$ is much larger than $ℓ$ ; for this reason in oscilloscopes the screen was placed far from the plates.

Exercise 4

An electron is launched with velocity $v_{0} = 4.0 \cdot 1 0^{6} m/s$ inclined at $4 5^{\circ}$ to the horizontal, in a region in which there is a uniform vertical field of magnitude $E = 500 N/C$ , directed upward. Compute the maximum height reached, the horizontal distance at which the electron returns to the starting height and the time taken.

a = \frac{e E}{m _{e}} (downward), h = \frac{v _{0}^{2} sin ^{2} 4 5 ^{\circ}}{2 a}, G = \frac{v _{0}^{2} sin 9 0 ^{\circ}}{a}, t_{v} = \frac{2 v _{0} sin 4 5 ^{\circ}}{a}

a h G t_{v} = \frac{1.60 \cdot 1 0 ^{- 19} \cdot 500}{9.11 \cdot 1 0 ^{- 31}} m/ s^{2} \approx 8.78 \cdot 1 0^{13} m/ s^{2} = \frac{( 4.0 \cdot 1 0 ^{6} ) ^{2} \cdot 0.50}{2 \cdot 8.78 \cdot 1 0 ^{13}} m \approx 4.6 \cdot 1 0^{- 2} m = \frac{( 4.0 \cdot 1 0 ^{6} ) ^{2}}{8.78 \cdot 1 0 ^{13}} m \approx 0.18 m = \frac{2 \cdot 4.0 \cdot 1 0 ^{6} \cdot 0.707}{8.78 \cdot 1 0 ^{13}} s \approx 6.4 \cdot 1 0^{- 8} s

The field points upward and the electron is negative, so the electric force points downward and the acceleration is constant, vertical, of magnitude $\frac{e E}{m _{e}}$ . The motion therefore has the same structure as projectile motion, with $a$ in place of $g$ : along the horizontal the velocity remains $v_{0} cos 4 5^{\circ}$ , along the vertical the motion is uniformly accelerated. The formulas for the maximum height, the range and the time of flight are obtained as in the topic on motion in the plane. The maximum height is about $4.6 cm$ and the electron returns to the starting height at a distance of about $18 cm$ , after $64 ns$ . The weight of the electron produces an acceleration about $1 0^{13}$ times smaller, so it is negligible.