7
Problem solving
Textbook: pp. 25–26
GoalSolve problems on electric field strength: charge, excess electrons, equilibrium conditions and finding ε.
New words
test charge · sinov zaryadiexcess electrons · ortiqcha elektronlarcondition of equilibrium · muvozanat shartidielectric constant · dielektrik singdiruvchanlik
Explanation
Field problems use three basic relations: E = F/q (so q = F/E), the field of a point charge E = k·|q|/(ε·r²), and q = N·e for a number of electrons. If ε is unknown, we find the vacuum value E₀ and then ε = E₀/E. A charged droplet hanging in air is in equilibrium: the upward electric force qE equals the weight mg, hence q = mg/E; in problems we take g = 10 m/s². If the droplet is negative, the field vector must point downward so that the force on it points up. Remember to convert mass to kilograms, distance to metres and charge to coulombs.
Worked examples
A force of 1.6·10⁻¹⁵ N acts on a body in a 500 N/C field. q = F : E = 1.6·10⁻¹⁵ : 500 = 3.2·10⁻¹⁸ C; the number of electrons is N = q : e = 3.2·10⁻¹⁸ : 1.6·10⁻¹⁹ = 20.
A droplet with charge q = 8 nC hangs in a field E = 2500 N/C. Its mass is m = q·E : g = 8·10⁻⁹ · 2500 : 10 = 2·10⁻⁶ kg = 2 mg.
Class activity
Make a two-column table in your notebook: on the left “given quantities”, on the right the matching formula (E = F/q, E = kq/r², q = Ne, mg = qE). Add one example of your own to every row.
Practice
1
In a liquid with ε = 2, at what distance from a 4 nC charge is the field 1800 N/C?
r² = k·q : (ε·E) = 9·10⁹ · 4·10⁻⁹ : (2 · 1800) = 0.01 m², r = 0.1 m = 10 cm.
2
What force in micronewtons acts on a 3 µC charge in a 200 N/C field?
600
3
A droplet of mass 1 mg carries 20 nC. What field strength in N/C keeps it hanging? (g = 10 m/s²)
500
4
If a hanging droplet is negative, why must the field vector point downward?
The force on a negative charge is opposite to E; for the force to point up (against the weight), E must point down.