Electricity

Electric Charge If you rub a glass rod, plastic ruler with a cloth, it will attract small pieces of paper. It is found that a rubbed glass rod repels a second charged glass rod when ones is brought close to the other. Similarly, if a rubbed ruler is brought close to another rubbed ruler, again a repulsive force is seen to act.

Electric Charge However, if the charged glass rod is brought close to the charged plastic ruler, it is found that they attract each other. Indeed, all charged objects fall into one of two categories. Either they are attracted or they are repelled.

Electric Charge Thus there seem to be two, and only two , types of electric charge. Each type of charge repels the same type, but attracts the opposite type. That is: Unlike charges attract; like charges repel. The two types of electric charge were referred to as positive and negative by Benjamin Franklin, who set the charge on the rubbed glass rod as positive charge, while the rubbed plastic ruler, negative.

Coulomb’s Law An electric charge exerts a force of attraction or repulsion on other electric charges. What factors affect the magnitude of this force? Charles Coulomb found that the force one tiny charged object exerts on a second tiny charged object is directly proportional to the charge on each of them. Charles Coulomb also found that the force decreased with the square of the distance between them.

Coulomb’s Law Coulomb’s Law Assume the magnitude of the charge of two objects are Q1, Q2, the distance between them is r, the force exerted on the objects is:

Coulomb’s Law Where k is a proportionality constant and has the value: The SI unit of charge is the coulomb (C). The charge on one electron has been determined to have a magnitude of 1.602×10-19C, and is negative. This is the smallest charge found in nature, and because it is fundamental, it is given the symbol e and is often referred to as the elementary charge:

Coulomb’s Law The constant k is often written in terms of another constant, , called the permittivity of free space. Coulomb’s law can be written as:

Coulomb’s Law Coulomb’s law looks a lot like the law of universal gravitation. Both are inverse square laws, both also have a proportionality to a property of each object, mass and charge. The major difference between the two laws is that gravity is always an attractive force, whereas the electric force can be either attractive or repulsive.

Coulomb’s Law Coulomb’s law can be written in vector form as: Where 𝑭 21 is vector force on charge 𝑄 2 due to 𝑄 1 , and 𝒓 12 is the unit vector pointing from 𝑄 1 to 𝑄 2 . If 𝑄 1 and 𝑄 2 have the same sign, the product 𝑄 1 𝑄 2 >0 and the force on 𝑄 2 away from 𝑄 1 , that is repulsive. If 𝑄 1 and 𝑄 2 have opposite sign, the product 𝑄 1 𝑄 2 <0 and the force 𝑭 21 point toward 𝑄 1 , that is attractive.

The Electric Field In particular, the electric field, 𝑬 , at any point in space is defined as the force 𝑭 exerted on a tiny positive test charge placed at that point divided by the magnitude of the test charge 𝑞 : 𝑬= 𝑭 𝑞 The reason for defining 𝑬 as 𝑭/𝑞 is that 𝑬 does not depend on the magnitude of the test charge 𝑞. That means 𝑬 describes only the effect of the charges creating the electric field at that point.

The Electric Field The electric field, for example, at a distance 𝑟 from a single point charge 𝑄 would have magnitude: 𝐸= 𝐹 𝑞 = 𝑘𝑄𝑞/ 𝑟 2 𝑞 𝐸=𝑘 𝑄 𝑟 2 = 1 4𝜋 𝜖 0 𝑄 𝑟 2 If the electric field 𝑬 is known at a given point in space, Then we can obtain the force 𝑭 on any charge 𝑞 placed At that point: 𝑭=𝑞𝑬 If q is positive, 𝑭 and 𝐸 point in the same direction. If 𝑞 is negative, 𝑭 and 𝑬 point in opposite directions.

The Electric Field In many cases we can treat charge as being distributed continuously. We can divide up a charge distribution into infinitesimal charges 𝑑𝑄, each of which will act as a tiny point charge. The contribution to the electric field at a distance r from each 𝑑𝑄 is: 𝑑𝑬= 1 4𝜋 𝜖 0 𝑑𝑄 𝑟 2 𝒆 𝒓

The Electric Field Then the electric field, 𝑬, at any point is obtained by summing over all the infinitesimal contributions, which is the integral: 𝑬= 𝑑𝑬 Note that 𝑑𝑬 is a vector, the integration should according to the vector’s Parallelogram law.

The Electric Field Example: A thin, ring-shaped object of radius 𝑎 holds a total charge +𝑄 distributed uniformly around it. Determine the electric field at a point P on its axis, a distance 𝑥 from the center. From symmetry of the object, we have: 𝑬= 𝑬 𝑥 So, 𝐸= 1 4𝜋 𝜖 0 𝑄 2𝜋𝑎 𝑑𝑙 𝑟 2 𝑐𝑜𝑠𝜃 𝐸= 1 4𝜋 𝜖 0 𝑄 𝑟 2 𝑐𝑜𝑠𝜃== 1 4𝜋 𝜖 0 𝑄𝑥 𝑥 2 + 𝑎 2 3 2

Gauss’s Law Electric Flux Gauss's law involves the concept of electric flux, which refers to the electric field passing through a given area. For a uniform electric field E passing through an area A, the electric flux Φ 𝐸 is defined as: Φ 𝐸 =𝑬⋅𝑨 Where 𝑨 is the vector whose magnitude is area A, and direction is perpendicular to the surface.

Gauss’s Law Example : Calculate the electric flux for (a) a spherical surface with radius 𝑅 (b) a cube with length 2𝑎, where the electric field is cause by a point charge Q at center. (a) the electric field is perpendicular to the spherical surface: Φ 𝐸 =4𝜋 𝑅 2 𝐸=4𝜋 𝑅 2 1 4𝜋 𝜖 0 Q R 2 Φ 𝐸 = 𝑄 𝜖 0

Gauss’s Law (b) For the symmetry of the cube, we should just calculate one surface and multiply by 6, so we have: Φ 𝐸 =6 −𝑎 𝑎 1 4𝜋 𝜖 0 𝑄𝑎 𝑥 2 + 𝑦 2 + 𝑎 2 3 2 𝑑𝑥𝑑𝑦 Φ 𝐸 =6 1 4𝜋 𝜖 0 𝑄 2 3 𝜋= 𝑄 𝜖 0

Gauss’s Law We can find out that the total electric flux doesn’t matter with the shape of the surface we integrated. The Gauss’s Law gives the precise relation between the electric flux through a closed surface and the net charge 𝑄 𝑒𝑛𝑐𝑙 enclosed within that surface: 𝑬⋅𝑑𝑨= 𝑄 𝑒𝑛𝑐𝑙 𝜖 0

Gauss’s Law Example: An electric charge 𝑄 is distributed uniformly throughout a nonconducting sphere of radius 𝑟 0 . Determine the electric field outside and inside the sphere. a) outside 4𝜋 𝑟 2 𝐸= 𝑄 𝜖 0 ⇔𝐸= 1 4𝜋 𝜖 0 𝑄 𝑟 2 b) Inside 4𝜋 𝑟 2 𝐸= 𝑄 𝜖 0 𝑟 3 𝑟 0 3 ⇔𝐸= 1 4𝜋 𝜖 0 𝑄𝑟 𝑟 0 3

Gauss’s Law Actually we can prove the Gauss’s law using Coulomb’s Law: If the charge density is 𝜌( 𝒓 ′ ), then the electric field is: 𝑬 𝒓 = 1 4𝜋 𝜖 0 𝜌 𝒓 ′ 𝒓− 𝒓 ′ 𝒓− 𝒓 ′ 3 𝑑 𝒓 ′ 𝛻⋅𝑬 𝒓 = 1 4𝜋 𝜖 0 𝜌 𝒓 ′ 𝛻⋅ 𝒓− 𝒓 ′ 𝒓− 𝒓 ′ 3 𝑑 𝒓 ′ = 1 4𝜋 𝜖 0 𝜌 𝒓 ′ 4𝜋 𝛿 3 (𝒓− 𝒓 ′ ) 𝑑 𝒓 ′ 𝛻⋅𝑬 𝒓 = 𝜌 𝒓 𝜖 0 𝛻⋅𝑬 𝑑 3 𝒓 = 𝑬⋅𝑑𝑨 = 𝜌 𝒓 𝜖 0 𝑑 3 𝒓 = 𝑄 𝑒𝑛𝑐𝑙 𝜖 0 𝑝𝑟𝑜𝑣𝑒𝑑!

Field Lines To visualize the electric field, we draw a series of lines to indicate the direction of the electric field at various points in space. These electric field lines are drawn so that they indicate the direction of the force due to the given field on a positive test charge.

Field lines An electric dipole is a combination of two equal charges of opposite sign. The electric field lines are curved and are directed from the positive charge to the negative charge. The direction of the electric field at any point is tangent to the field line at that point as shown by the vector arrow 𝑬 at point P.

Conductors and Insulators Suppose we have two metal spheres, one highly charged and the other electrically neutral. If we place a metal object to touch both spheres, the previously unchanged sphere quickly becomes charged. If, instead, we connect the two spheres by a wood rod, the uncharged ball will not become charged. Metals are said to be conductors of electricity, whereas wood is an insulator.

Conductors and Insulators In the atomic point of view, the electrons in an insulating material are bound very tightly to the nuclei, while in the conductors there are some electrons are bound very loosely, so called free electrons. The free electrons can move swiftly away from a negatively charged object or toward a positively charged object. It is the free electrons which give metal the conductivity.

Motion of a Charged Particle in an Electric Field If an object having an electric charge 𝑞 is at a point in space where the electric field is 𝐸, the force on the object is given by: 𝑭=𝑞𝑬 Mostly this force is much greater than the gravitational force, if 𝑞𝐸≫𝑚𝑔, the gravitational force can be ignored. For a single electron, this condition is easily to reach: 𝐸≫ 𝑚𝑔 𝑞 = 9.1× 10 −31 𝑘𝑔×9.8𝑚/ 𝑠 2 1.6× 10 −19 𝐶 ≈5.6× 10 −11 N/C

Motion of a Charged Particle in an Electric Field Example: The electric field, 𝐸= 10 −4 𝑁/𝐶, the distance, 𝑑=1𝑚, the initial speed 𝑣 0 =1000𝑚/𝑠, and is perpendicular to the electric field. Calculate the vertical distance L the electron travels.

Motion of a Charged Particle in an Electric Field Solution: The acceleration is: 𝑎 𝑥 = 𝑞𝐸 𝑚 =−𝑒𝐸/𝑚, the motion of electron is projectile motion: 𝑥=−𝑑= 1 2 𝑎 𝑥 𝑡 2 =− 𝑒𝐸 2𝑚 𝑡 2 𝑡= 2𝑚𝑑 𝑒𝐸 So 𝐿= 𝑣 0 𝑡= 𝑣 0 2𝑚𝑑 𝑒𝐸 =1000× 2×9.1× 10 −31 ×1 1.6× 10 −19 × 10 −4 𝑚=0.337𝑚

Electric Dipoles The combination of two equal charges of opposite sign +𝑄 and −𝑄, separated by a distance 𝑙, is referred to as an electric dipole. The quantity 𝑄𝒍 is called the dipole moment 𝒑, and it points from negative to the positive charge.

Electric Dipoles Many molecules, such as the water molecules, have a dipole moment. Even though the molecule as a whole is neutral, there is a separation of charge that results from an uneven sharing of electrons by the two atoms. In the water molecule ( 𝐻 2 𝑂), the electrons spend more time around the oxygen atom than around the hydrogen atoms. The net dipole moment 𝒑 can be considered as the vector sum of two dipole moments 𝒑 𝟏 and 𝒑 𝟐 that point from the O toward each H, as shown:

Electric Dipoles Let us consider a dipole, the dipole moment 𝒑=𝑄𝒍, that is placed in a uniform electric field 𝑬. We found that there is no net force on the dipole, but a torque on it, which has magnitude: 𝜏= 𝜏 + + 𝜏 − =𝑄𝐸 𝑙 2 𝑠𝑖𝑛𝜃+𝑄𝐸 𝑙 2 𝑠𝑖𝑛𝜃 This can be written in vector notation as: 𝝉=𝒑×𝑬

Electric Dipoles The effect of torque is to try to turn the dipole so 𝒑 is parallel to 𝑬. The work done on the dipole by the electric field to change the angle 𝜃 from 𝜃 1 to 𝜃 2 is: 𝑊= 𝜃 1 𝜃 2 𝜏𝑑𝜃 =−𝑝𝐸 𝜃 1 𝜃 2 𝑠𝑖𝑛𝜃𝑑𝜃 =𝑝𝐸(𝑐𝑜𝑠 𝜃 2 −𝑐𝑜𝑠 𝜃 1 )

Electric Dipoles Positive work done by the field decreases the potential energy, 𝑈, of the dipole in this field. If we choose 𝑈=0 when 𝒑 is perpendicular to 𝑬, and setting 𝜃 2 =𝜃, then: 𝑈=−𝑊=−𝑝𝐸𝑐𝑜𝑠𝜃=−𝒑⋅𝑬