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New YorkPhysicsSyllabus dot point

How do objects become charged, and how does the force between charges depend on charge and distance?

Describe charging by friction, conduction and induction, state that charge is conserved and quantised in multiples of the elementary charge, and apply Coulomb's law Fe=kq1q2/r2F_e = kq_1q_2/r^2 to calculate the force between point charges.

A Regents Physics answer on static electricity and Coulomb's law: how objects are charged by friction, conduction and induction, the conservation and quantisation of charge, and how to apply the Reference-Table equation for the force between point charges, with worked examples.

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  1. What this topic is asking
  2. Charging by friction, conduction and induction
  3. Conservation and quantisation of charge
  4. Coulomb's law
  5. Reference Tables note
  6. Try this

What this topic is asking

This dot point opens the electricity strand with static electricity: how objects gain charge, the rules charge obeys, and the Coulomb's law force between charges. The Physical Setting/Physics course asks you to describe charging by friction, conduction and induction, to state that charge is conserved and comes in whole multiples of the elementary charge, and to calculate the electrostatic force with Fe=kq1q2r2F_e = \dfrac{kq_1 q_2}{r^2}. The Regents tests charging processes conceptually and Coulomb's law as a calculation.

Charging by friction, conduction and induction

Only electrons move in solids; the positive nuclei stay fixed. So an object becomes negative by gaining electrons and positive by losing them. Rubbing a balloon on hair transfers electrons to the balloon (friction); touching a charged rod to a sphere shares the charge (conduction); bringing a charged rod near a neutral conductor pushes its electrons to the far side, which can then be grounded to leave a net charge (induction). Induction is the only method that charges an object opposite to the inducing charge without contact.

Conservation and quantisation of charge

Conservation means that if one object gains a certain negative charge, another loses exactly that much (gains positive charge). Quantisation means a charge of, say, 2.4×10−192.4 \times 10^{-19} C is impossible for a single particle, since it is not a whole multiple of ee. Counting electrons transferred and multiplying by ee is a standard Regents calculation.

Coulomb's law

Coulomb's law mirrors the law of gravitation: both are inverse-square laws, so doubling the distance quarters the force. The differences are that the electrostatic force is enormously stronger (because kk is huge while GG is tiny) and that it can be attractive or repulsive, whereas gravity is only attractive. The same inverse-square reasoning applies: square the distance factor to scale the force.

Reference Tables note

Coulomb's law Fe=kq1q2r2F_e = \dfrac{kq_1 q_2}{r^2} is printed in the Electricity section of the Reference Tables, and the constants k=8.99×109k = 8.99 \times 10^9 N m squared per C squared and e=1.60×10−19e = 1.60 \times 10^{-19} C are in the constants list. The charging processes (friction, conduction, induction) and the conservation and quantisation of charge are stated principles you recall. The link from force to field, E=FeqE = \dfrac{F_e}{q}, leads into electric fields and potential.

Try this

Q1. State how an object becomes negatively charged in terms of electrons. [1 point]

  • Cue. It gains electrons (only electrons move in solids; gaining them gives a net negative charge).

Q2. State what happens to the Coulomb force between two charges if the distance between them is halved. [1 point]

  • Cue. It becomes four times as large (inverse-square: halving rr multiplies the force by 22=42^2 = 4).

Exam-style practice questions

Practice questions written in the style of NYSED exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Regents (style)2 marksPart B-2 (constructed response). Two point charges of +3.0×10−6+3.0 \times 10^{-6} C and +5.0×10−6+5.0 \times 10^{-6} C are placed 0.200.20 m apart. Using k=8.99×109k = 8.99 \times 10^9 N m squared per C squared, calculate the magnitude of the electrostatic force between them. Show the equation, substitution and answer.
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A 2-point constructed-response calculation using the Reference-Table equation Fe=kq1q2r2F_e = \dfrac{kq_1 q_2}{r^2}.

Equation: Fe=kq1q2r2F_e = \dfrac{kq_1 q_2}{r^2}.
Substitution: Fe=(8.99×109)(3.0×10−6)(5.0×10−6)(0.20)2F_e = \dfrac{(8.99 \times 10^9)(3.0 \times 10^{-6})(5.0 \times 10^{-6})}{(0.20)^2}.
Answer: Fe=(8.99×109)(1.5×10−11)0.040=3.4F_e = \dfrac{(8.99 \times 10^9)(1.5 \times 10^{-11})}{0.040} = 3.4 N, repulsive (both charges positive).

Markers reward the equation from the tables, correct substitution including the squared distance, and noting the force is repulsive. A common error is omitting the square on the distance.

Regents (style)1 marksPart A (multiple choice). A neutral metal sphere gains 5.0×10125.0 \times 10^{12} electrons. What is the sign and approximate magnitude of its charge (e=1.60×10−19e = 1.60 \times 10^{-19} C)? (1) positive, 0.800.80 C (2) negative, 0.80×10−60.80 \times 10^{-6} C (3) negative, 1.0×10−61.0 \times 10^{-6} C (4) positive, 1.0×10−61.0 \times 10^{-6} C. Justify your reasoning.
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A 1-point Part A item on quantisation of charge. The answer is (2).

Gaining electrons makes the sphere negative. The magnitude is the number of electrons times the elementary charge: q=(5.0×1012)(1.60×10−19)=8.0×10−7q = (5.0 \times 10^{12})(1.60 \times 10^{-19}) = 8.0 \times 10^{-7} C, or 0.80×10−60.80 \times 10^{-6} C. The trap is the sign: adding electrons gives a negative charge, not positive.

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