Lessons 33 · 1 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 10, 1st edition. Niso Poligraf Publishing House, Tashkent, 2017
33
Work done in moving a charge in an electrostatic field
Textbook: pp. 123–124
GoalCalculate the work done in moving a charge in an electrostatic field (A = qEd, A = q(φ₁ – φ₂)), its independence of the path and the acceleration of a charged particle.
New words
electrostatic field · elektrostatik maydonpotential field · potensial maydonconservative force · konservativ kuchelectronvolt · elektronvolt
Explanation
In a uniform field, moving a charge q a distance d along a field line the field does work A = qEd; if the angle between the force and the displacement is α, A = qEs·cosα = qEΔx, where Δx is the projection of the displacement on the field direction. The work depends not on the shape of the path but only on the initial and final points: the electrostatic field is a potential field and its force is conservative; around a closed loop the work is zero. In terms of potentials: A = q(φ₁ – φ₂) = qU = –ΔWₚ. If a charged particle is accelerated from rest through a voltage U, the field’s work becomes kinetic energy: qU = mυ²/2, so υ = √(2qU/m). In atomic physics energy is measured in electronvolts: 1 eV = 1.6·10⁻¹⁹ J (the energy gained by the elementary charge e = 1.6·10⁻¹⁹ C across 1 V). The mass of an electron is 9.1·10⁻³¹ kg, of a proton 1.67·10⁻²⁷ kg.
Worked examples
In a uniform field E = 5000 V/m a charge q = 3 nC moves 0.2 m along the field: A = qEd = 3·10⁻⁹·5000·0.2 = 3·10⁻⁶ J = 3 μJ. If the displacement of the same length makes 60° with the field (q = 4 nC): A = 4·10⁻⁹·5000·0.2·0.5 = 2 μJ.
An electron accelerated from rest through U = 100 V gains 100 eV = 1.6·10⁻¹⁷ J; υ = √(2qU/m) = √(2·1.6·10⁻¹⁷/9.1·10⁻³¹) ≈ 5.9·10⁶ m/s.
Class activity
Only in class, by calculation: in your notebook justify with the formula why a proton and an electron accelerated through the same voltage have equal kinetic energies but different speeds because of their masses. No experiment is done; high voltage is dangerous.
Practice
1
What is the work around a closed loop in an electrostatic field, and why?
Zero: the work depends only on the initial and final points (the field is potential).
2
What is the work of moving q = 2 nC 0.5 m along a field of E = 1000 V/m (nJ)?
1000
3
A 5 nC charge moves from a point at 80 V to one at 20 V. What is the field’s work (nJ)?
300
4
Why is the field’s work zero for a displacement perpendicular to the field lines?
cos90° = 0 (along an equipotential surface the potential difference is zero).