Electric Field, Coulomb Law and Charge Calculations
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Electric Field, Force and Charge Calculations
Definitions and Basic Characteristics
Charge E is used only for electric charges and is a vector quantity.
To check the presence of the field E, observe the electrostatic forces produced when there are two charges.
Two aspects affecting acceleration
- Mass: larger mass leads to smaller acceleration (for the same force).
- Force: acceleration follows Newton's 2nd law, a = F / m.
Basic characteristic of charge
Basic charge values: the proton has charge +1.6 × 10-19 C; the electron has charge -1.6 × 10-19 C. These values are fixed for these particles. An atom can gain or lose electrical charge by gaining or losing electrons, becoming a charged particle called an ion.
Key Formulas
Force on a charge in an electric field: F = q × E.
Acceleration from force: a = F / m = qE / m.
Electric field of a point charge: E = k × q / r2.
Notation: e = electric field, F = force, q = charge, a = acceleration, m = mass, r = distance. Coulomb's constant: k = 9 × 109 N·m2/C2.
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Problem 2: Alpha Particle in a Uniform Electric Field
Determine the acceleration of an alpha particle which moves in a uniform electric field. (Original data transcription follows; corrected for capitalization and grammar but content preserved.)
Q: Tell relation ideas between it and an experiment by the proton. Data: a = / / 32x10-2C x 1w / c / / 668x10-29 Kg
Result shown: a = 0.04 x 109 m/s2 → a? = 4 x 107 m/s2
Problem 3: Two Electron Charges Separated by 10 cm
Three cases are presented. Two charges q1 and q2 with values 2 × 10-2 C and 3 × 10-2 C are spaced 10 cm apart. Calculate the resulting electric field E in the indicated cases. Use k = 9 × 109 N·m2/C2. Consider the point asked to calculate: there is a positive charge.
Case A
Configuration (on line):
5 cm (2 × 10-2 C) _____ 5 cm _____ 3 × 10-2 C
Given: q1 = 2 × 10-2 C, q2 = 3 × 10-2 C, r1 = r2 = 5 cm = 5 × 10-2 m.
Development:
E1 = k × q2 / r12 = 9 × 109 N·m2/C2 × 2 × 10-2 C / (5 × 10-2 m)2 = 18 × 107 N·m2/C / 25 × 10-4 m2 = 0.7 × 1011 N/C.
k × q2 / r22 = E2 = the same but with 3 (i.e., scaled for 3 × 10-2 C).
Er = E1 - E2 = 0.7 × 1011 N/C - 1.08 × 1011 N/C = -0.38 × 1011 N/C.
Case B
Configuration: 4 cm 2 × 10-2 C +6 cm 3 × 10-1 C. Direction: horizontal. Result (magnitude): 0.4 × 1011 N/C.
Case C
Configuration: 4 cm 2 × 10-2 C - - - +6 cm 3 × 10-2 C. Direction: horizontal (west). Magnitude: 1.2 × 1011 N/C.
Lines of Force
What are the lines of force? They are imaginary lines tangent to the direction of the electric field at each point.
- Like charges repel: the lines of force move away from each other, going to opposite sides.
- Opposite charges attract: lines of force point from positive charges toward negative charges.
Example: Particle Acceleration in an Electric Field
2) A particle undergoes acceleration k (given) if its charge is q and it moves in a field E of 0.7 N/C and has mass 2.8 × 10-4 kg.
Data: q = 8e → q = 12.8 × 10-19 = 128 × 10-20 C. E = 0.7 N/C = 7 × 10-1 N/C. m = 2.8 × 10-27 kg (as transcribed).
Development: a = qE / m = (128 × 10-20 C × 7 × 10-1 N/C) / (2.8 × 10-27 kg) = ... (transcribed result) = a = 32 × 107 N/kg.
Problem: Find Charge from a Given Field
3) A field of 5 × 10-2 N/C is created by a charge Q located 20 cm from a point. Determine the value of Q in coulombs (C) and in STC (as transcribed).
Data: E = 5 × 10-2 N/C, r = 20 cm = 2 × 10-1 m.
Development: E = k × Q / r2 → Q = E r2 / k.
Compute: 5 × 10-2 N/C × (2 × 10-1 m)2 / (9 × 109 N·m2/C2) = (5 × 10-2 × 4 × 10-2) / 9 × 109 = 20 × 10-4 / 9 × 109 → Q ≈ 2.2 × 10-12 C.
As transcribed: 3 x 109 = STC. Q = 6.6 x 10-3 STC (transcribed notation preserved).
Note: The original document content and numeric expressions have been preserved and reformatted for clarity. Spelling, punctuation, capitalization and unit notation were corrected while retaining the original sequence of statements and computed results as presented.