9478 / 2027
Electric Fields overview

Topic 5 of 7

Charged-particle motion

Calculate the force first, then resolve the motion. In a uniform field, acceleration is constant, but a component perpendicular to the field keeps its velocity if no other force acts.

Connect force to components

For constant mass with electric force alone, a = qE/m as a vector relation. A positive charge initially moving along E speeds up. A negative charge initially moving along E slows and may reverse if it remains in the field. For entry perpendicular to E, the initial velocity component along E is zero; that component then changes under the electric force, while the velocity component perpendicular to E stays constant.

Use the classical particle model below with a fixed electrostatic field. Neglect gravity, radiation and edge/fringing effects. Outside the specified ideal field region, assume no other force.

Worked electron deflection

Calculate transit time from the horizontal motion

An electron enters horizontally at the midpoint between plates separated by 0.0300 m. The upper plate is at +120 V and the lower at 0 V, giving a downward field of 4000 V/m. The plates are 0.0600 m long and the initial horizontal speed is 2.00 × 107 m/s.

Take x rightward and y upward, with entry at x = y = 0. Use q = -1.60 × 10-19 C and electron mass me = 9.11 × 10-31 kg.

Fy = +6.40 × 10-16 N
ay = Fy/me ≈ +7.03 × 1014 m/s2
ax = 0; vx = ux
texit = 0.0600/(2.00 × 107)
= 3.00 × 10-9 s = 3.00 ns

One nanosecond is 10-9 s. With initial vy = 0, the component equations are:

x = uxt
y = (1/2)ayt2
vy = ayt
Calculated trajectory coordinates for the supplied ideal field
t / nsx / my / mm
000
1.000.02000.351
2.000.04001.405
3.000.06003.161

The electron accelerates upward while moving right

Use the same 0.0300 m plate gap and 4000 V/m downward field, a plate length of 0.0600 m and horizontal entry speed 2.00 × 107 m/s. Neglect gravity, fringing and radiation in this classical model. Upward y is positive.

Retain the actual small deflection

The electron exits 3.16 millimetres above its midpoint entry and then travels along the tangentThe physical x and y distance scales are equal, at four thousand drawing units per metre. The plates are 240 units long and 120 units apart, representing 0.0600 and 0.0300 metres. Entry is at their midpoint, x zero and y zero. Filled dots show alternative positions of the same electron at zero, one, two and three nanoseconds. Their physical x values are zero, 0.0200, 0.0400 and 0.0600 metres; upward displacements are zero, 0.351262, 1.405049 and 3.161361 millimetres. The blue parabolic path is generated from constant upward acceleration 7.02524698 times ten to the fourteen metres per second squared. At the two-nanosecond point, a purple force arrow points upward, while the separate teal field arrows point downward. The initial horizontal velocity vector is shown separately above the plates, in blue and labelled initial u sub x; arrow lengths are not a common scale for these different quantities. The field boundary is at the plate end. A dashed straight continuation follows the actual exit tangent at 6.01555 degrees above horizontal in the force-free ideal exterior. No vertical magnification exaggerates the deflection. The lower coordinate ruler measures x from entry; the vertical gap bracket measures the physical plate separation.Equal physical x and y scalesDots: 0, 1, 2 and 3 nsInitial uxUpper plate: +120 VLower plate: 0 VE = 0y = 0F30.0 mm gap+y0.0000.0200.0400.060x / m

The plate gap and trajectory use one distance scale: the exit rise is only 3.16 mm, safely below the 15.0 mm half-gap. Blue is velocity/path direction, teal is E and purple is the force on the electron. The dashed continuation is straight because the ideal field has ended.

Horizontal velocity stays constant

Horizontal velocity stays constant during the three-nanosecond transitTime is zero to three nanoseconds on the same horizontal scale as the other component plot, eighty drawing units per nanosecond. The vertical quantity is horizontal velocity v sub x in millions of metres per second, with ticks from zero to twenty-five. The line stays at twenty throughout because there is no horizontal force. The two component plots use different labelled vertical scales; compare values and units rather than visual heights. They end at the field exit, not after the later straight-path interval.01230510152025vx / 106 m s-1t / ns

Upward velocity increases linearly

Upward velocity increases linearly during the three-nanosecond transitTime is zero to three nanoseconds on the same horizontal scale as the other component plot, eighty drawing units per nanosecond. The vertical quantity is upward velocity v sub y in millions of metres per second, with ticks from zero to 2.5. The straight line starts at zero and ends at 2.10757409 at three nanoseconds. Its slope represents the constant upward acceleration 7.02524698 times ten to the fourteen metres per second squared. The two component plots use different labelled vertical scales; compare values and units rather than visual heights. They end at the field exit, not after the later straight-path interval.012300.511.522.5vy / 106 m s-1t / ns

At exit, vx = 20.0 × 106 m/s and vy ≈ 2.11 × 106 m/s. The exit direction is about 6.02° above horizontal. A short time in the field gives a small change in direction, even though the acceleration is large.

The trajectory uses equal physical horizontal and vertical scales, retaining the small deflection. Beyond the plate region the continuation is tangent to the path. The two velocity-time graphs share their time interval but label different vertical scales.

Keeping extra digits in the acceleration until the final result gives:

yexit ≈ +3.16 × 10-3 m = +3.16 mm
vy,exit ≈ +2.11 × 106 m/s
tan θ = vy,exit/vx
θ ≈ 6.02° above horizontal

The upward displacement is less than the 15.0 mm upper half-gap, so the electron exits without hitting the upper plate in this model. The chosen speed is treated with the stated approximate classical equations.

Inside the field, eliminating t gives y = ayx2/(2ux2), a parabola. After exit, acceleration is zero and both velocity components keep their exit values. The electron follows a straight line along the exit tangent; earlier acceleration does not keep bending it after the force ends.

Optional check An electron enters the midpoint of the plate field horizontally at 2.00 x 10^7 m/s. Its upward acceleration is 7.025 x 10^14 m/s^2 over the 0.0600 m plate length. What happens at and after exit in the stated force-free exterior?
An electron enters the midpoint of the plate field horizontally at 2.00 x 10^7 m/s. Its upward acceleration is 7.025 x 10^14 m/s^2 over the 0.0600 m plate length. What happens at and after exit in the stated force-free exterior?

Check the motion with energy

Potential increases upward with gradient 4000 V/m. For the calculated exit displacement:

ΔV = 4000yexit ≈ +12.6 V
ΔUE = qΔV ≈ -2.02 × 10-18 J
ΔEk = -ΔUE ≈ +2.02 × 10-18 J

The negative charge gains kinetic energy while moving to higher potential. Since its horizontal velocity is unchanged, the same gain is (1/2)mevy,exit2. Potential, potential energy and charge are distinct: the electron's charge has not changed.

Worked acceleration from rest

An electron moves to a potential 150 V higher

In a separate field-only example, let ΔV = +150 V and q = -e, where e = 1.60 × 10-19 C is the positive elementary-charge magnitude. With negligible other transfers:

ΔEk = -qΔV = e(150 V)
= 2.40 × 10-17 J
v = √(2ΔEk/me)
≈ 7.26 × 106 m/s

The initial kinetic energy is zero in this example. That is a different initial condition from the horizontally entering electron above.