Electrical Charges and Coulomb’s LawPhys 2180 Lecture (1)

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Structure of Matter

• Fundamental building blocks of the matter are atoms.

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Structure of Matter

• Neutral atom – electron = Positive ion

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C101.602charge electron -191

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Structure of Matter

• Neutral atom + electron = negative ion.

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Electrical Charges

• Electrostatics is the study of electric charge at rest. • Electric charge is a fundamental property of matter.

– Two types of electric charges• Positive charge - every proton has a single positive charge.• Negative charge - every electron has a single negative

charge.

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Fundamental Charges

Note that the electron and proton both have the same charge, with the electron being negative and the proton being positive. This amount of charge is often called the

electronic charge, e. This electronic charge is generally considered a positive value (just like g in gravity). We add the negative sign when we need to:

qe = -e; qp = +e.

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•Electrostatics is the study of electric charge at rest.

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Electric chargeTwo different kinds of charge, positive and negative.Electricity, with attractive and repulsive forces, needs two kinds of charge.

• Electric charge is measured in coulombs (C)

Properties of electric charge

1.Two kinds of charges occur in nature, o like charges repel one another, o and unlike charges attract one another.2. Charge is conserved.3. Charge is quantized.

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• An object becomes electrostatically charged by:

1. Friction,which transfers electrons between two objects in contact,

2. Contact with a charged body which results in the transfer of electrons,

3. Induction which produces a charge redistribution of electrons in a material.

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Types Of Forces• There are only four fundamental forces of nature.

• Gravitational Force

• Electromagnetic Force

• Strong Nuclear Force

• Weak Nuclear Force10

Electric Field - Definition

• the electric field vector E at a point in space is defined as the electric force Fe acting on a positive test charge q placed at that point divided by the test charge:

Note that since F is a vector and q is a scalar, E must be a vector.the units of Electric Field in SI units of newtons per coulomb (N/C)

F = K|q||q|/r2 = magnitude of the electric forcek = Coulomb's constant = 8.9875 x 109Nm2/c2 

E = K|q| / r2

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Electric Field Lines

• The electric field lines for a point charge.

• (a) For a positive point charge, the lines are directed radially outward.

• (b) For a negative point charge, the lines are directed radially inward.

• Or the electric field lines extend away from positive charge (where theyoriginate) and towards negative charge (where they terminate) 12

Electric Force: Coulomb’s Law

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Electric Force: Coulomb’s Law• we can express Coulomb’s law as an equation giving the

magnitude of• the electric force (sometimes called the Coulomb force)

between two point charges:

where ke is a constant called the Coulomb constant, q1 and q2 the charges, r is the distance between the charges.

We also found that the force decreases with distance between the charges just like gravity.

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Electric Force: Coulomb’s Law

The value of the Coulomb constant depends on the choice of units. The SI unit of charge is the coulomb (C).

The Coulomb constant ke in SI units has the value

where the constant epsilon is known as the permittivity of free space ε0  (8.854 187 817)*10≅ -12 C2/(N m2)

151 Coulomb = 106 microCoulomb 1 Coulomb = 109 nanoCoulomb

Problem (1)

• The average distance r between the electron and the proton in the hydrogen atom is 5.3x10-11 m. (a) What is the magnitude of the average electrostatic force that acts between these two particles? (b) What is the magnitude of the average gravitational force that acts between these particles?

• Solution • the electrostatic force,

• the gravitational force,

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= = 8.2 X 10-8 N

= = 1.91 X 10-57 N

• The nucleus of an iron atom has a radius of about 4x10-15 m and contains 26 protons. What repulsive electrostatic force acts between two protons in such a nucleus if a distance of one radius separates them?

• Solution

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Problem (2)

= = 14 N

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Problem (3)

Two balloons with charges of +3.37 µC and -8.21 µC attract each other with a force of 0.0626 N. Determine the separation distance between the two balloons.

Solution:Given:q1 = +3.37 µC = +3.37 x 10-6 C , k= q2 = -8.21 µC = -8.21 x 10-6 CFe = -0.0626 N (use a - force value since it is attractive)

Find: r = ???

𝐹 𝑒=𝐾𝑞1𝑞2

𝑟2

𝑟2=𝐾𝑞1𝑞2

𝐹 𝑒

r

¿√( 8 . 99   x  109 Nm 2/ C 2 )  ( −8 .21  x  10 −6  C )(+3 .37  x  10 −6 C ) ( −0 . 0626  N )

r =1.99 m

Three Charges on a Line

Determine the magnitude and direction of the net force on q1.

Problem (4)

N7.2m20.0

C100.4C100.3CmN1099.82

66229

221

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r

qqkF

N4.8m15.0

C100.7C100.3CmN1099.82

66229

231

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r

qqkF

5.7NN4.8N7.21312 FFF

Problem (4)

18.5 Coulomb’s Law Find the net force on q1

cossin 1312 FFF

18.5 Coulomb’s Law

N59.9m15.0

C100.6C100.4CmN1099.82

2626229

2

2112

r

qqkF

N98.17m10.0

C100.5C100.4CmN1099.82

2626229

231

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r

qqkF

NFFF 43.1473cos98.1773sin59.9cossin 1312