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Electromagnetism overview

Topic 1 of 7

Motion in a uniform electric field

An electric field exerts a force according to charge sign. In a uniform field, that force has a fixed direction, so resolve the motion along and perpendicular to the field.

Electric field direction is the force direction on a positive test charge. For a particle with signed charge Q, the vector relation is F = QE. Its force magnitude is |Q|E: positive charge is forced along the field and negative charge against it. Field strength E has unit N/C, equivalent to V/m.

With unchanged charge and mass and no other significant force, Newton's second law gives constant acceleration a = QE/m in a chosen direction. The electric force exists even if the particle is initially stationary. Its direction does not depend on the particle's initial velocity.

Parallel and perpendicular entry

A positive particle already moving along E speeds up. Moving against E, it slows and can reverse. A negative particle moving along E instead slows and can reverse, because its force is opposite E; moving against E, it speeds up. State both the charge sign and the velocity direction before deciding.

With initial velocity perpendicular to E, the component perpendicular to the field remains constant while the component along the field changes. Constant motion in one direction combined with constant acceleration in the other produces a parabola. Use the constant-acceleration equations separately for the two components.

Worked electron beam

Find the complete exit state

An electron enters horizontally right at 2.00 × 107 m/s into a uniform field 4000 N/C downward. The field region is 0.0600 m long and the vertical plate separation is 0.0300 m. Entry is halfway between the plates. The upper plate is positive relative to the lower.

Use electron charge Q = -e, with e = 1.60 × 10-19 C, and electron mass me = 9.11 × 10-31 kg. Neglect gravity, collisions, fringing and radiation, and use the classical motion model. The field strength is supplied directly.

A small deflection at equal position scales

A small deflection at equal position scalesThe plate edges run from x sixty to three hundred drawing units, with upper edge at one hundred and sixty and lower at two hundred and eighty. Four thousand drawing units represent one metre on both physical axes. The electron enters at sixty, two hundred and twenty and leaves at three hundred, 207.354555, so its upward displacement is 3.161361 millimetres. Filled points at one, two and three nanoseconds lie on the exact parabola. The electric field points down and the electron force up. Outside the right edge, the trajectory continues on its true exit tangent. Force and field arrows have separate schematic lengths, not position or common magnitude scales.Given E = 4000 N/C down+ + + + +- - - - -Ev inFField length: 0.0600 mPlate gap: 0.0300 m

Positions use the same scale horizontally and vertically. The electron remains inside the 0.0300 m gap. Its horizontal speed stays constant while its total speed increases; the straight continuation is the exit tangent.

The electron is forced upward in the downward electric field. Equal horizontal and vertical distance scales preserve the small deflection; the path becomes a straight exit tangent outside the ideal field region. Field and force arrows represent different quantities.

Take right and up as positive. The field component is Ey = -4000 N/C, so the negative charge gives an upward force:

Fy = QEy
= (-1.60 × 10-19)(-4000)
= +6.40 × 10-16 N
ay = Fy/me
≈ +7.025 × 1014 m/s2

An acceleration of order 1015 m/s2 acting for a few nanoseconds suggests a displacement on a millimetre scale. The short transit time matters even though the acceleration is very large.

There is no horizontal force, so vx remains 2.00 × 107 m/s. Initially vy = 0, but the total initial velocity is not zero:

t = 0.0600/(2.00 × 107)
= 3.00 × 10-9 s = 3.00 ns
y = ½ayt2 = +0.003161 m
vy = ayt = +2.108 × 106 m/s
Calculated states of the supplied electric-beam model; vx is constant
t / nsy / mmvy / (106 m/s)
000
1.000.35130.7025
2.001.4051.405
3.003.1612.108

The exit displacement is 3.16 mm upward, less than the 15.0 mm half-gap, so the electron reaches the far edge without striking the upper plate. Its total exit speed is:

v = √(vx2 + vy2)
≈ 2.011 × 107 m/s
Exit angle = tan-1(vy/vx) ≈ 6.02° above right

Keeping extra digits makes the speed increase visible; to three significant figures it is 2.01 × 107 m/s. The field has done positive work on the electron. Constant horizontal velocity does not imply constant total speed.

Outside the ideal field, with no further force, the electron keeps its exit velocity and follows a straight tangent. Reversing the charge sign reverses the force for the same E; a quantitative path comparison also needs the particle's mass and entry velocity.

Optional check An electron enters a uniform downward electric field moving horizontally right. Neglect other forces. While it is inside the field, which description is correct?
An electron enters a uniform downward electric field moving horizontally right. Neglect other forces. While it is inside the field, which description is correct?