Work Done in Electric Fields: Understanding Energy Transfer

Expert reviewed • 23 November 2024 • 5 minute read

Key Concepts

Electric Field and Work

In a uniform electric field, the work done (W) on a charged particle depends on several factors:

  • The magnitude of the charge (q)
  • The electric field strength (E)
  • The displacement (d)
  • The potential difference (V)

The relationships between these quantities are expressed through these equivalent equations:

W=qVW = qV W=qEdW = qEd

Where:

  • W is work done in joules (J)
  • q is charge in coulombs (C)
  • V is potential difference in volts (V)
  • E is electric field strength in newtons per coulomb (N/C)
  • d is displacement in meters (m)

Kinetic Energy Relationship

The work done on a charged particle changes its kinetic energy according to:

W=ΔK=12mv2W = \Delta K = \frac{1}{2}mv^2

Where:

  • m is the mass of the particle in kilograms (kg)
  • v is the velocity in meters per second (m/s)

Analyzing Particle Motion

When a charged particle enters a uniform electric field:

  • It experiences a constant force F = qE
  • This results in constant acceleration a = qE/m
  • The particle follows a parabolic path (if entering at an angle)
  • The work done depends only on the total displacement in the direction of the field

Practice Question 1

An electron (q = -1.60 × 10⁻¹⁹ C) enters a uniform electric field at 50 m/s between parallel plates separated by 2 cm with a potential difference of 7 V. placeholder
a) Electric field strength: $$E = \frac{V}{d} = \frac{7\text{ V}}{0.02\text{ m}} = 350\text{ N/C}$$

b) Electron acceleration: a=qEm=(−1.60×10−19)(3509.11×10−31=−6.15×1013 m/s²a = \frac{qE}{m} \\= \frac{(-1.60 × 10^{-19})(350}{9.11 × 10^{-31}} \\= -6.15 × 10^{13}\text{ m/s²}

c) Work done to move electron from positive to negative plate: W=qV=(−1.60×10−19)(7)=−1.12×10−18 JW = qV \\= (-1.60 × 10^{-19})(7) \\= -1.12 × 10^{-18}\text{ J}

Summary

  • Work done in electric fields can be calculated using either W=qVW = qV or W=qEdW = qEd
  • The work done changes the particle's kinetic energy
  • Understanding these relationships is crucial for analyzing particle behavior in electric fields

Practice Problems

  • Calculate the work done when a proton moves through a potential difference of 100 V.
  • Determine the final velocity of an electron accelerated from rest through 5 kV.
  • Find the electric field strength between plates separated by 5 cm with a potential difference of 1000 V.

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Motion of Charged Particles in Electric Fields

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Magnetic Forces on Moving Charges

Return to Module 6: Electromagnetism