Lecture 2 : Electric charges and fields Electric field Introducing Gauss’s Law

Electric charge Electric charge is an intrinsic property of the particles which make up matter, which experimentally can be either positive or negative

Electric charge Electric charge is quantized such that protons and electrons have equal and opposite charge 𝒆=±𝟏.𝟔× 𝟏𝟎 −𝟏𝟗 𝑪 [unit 𝐶= Coulombs]

Electric charge Electric charge is locally conserved and cannot be created or destroyed

Electric charge Two electric charges attract or repel each other with equal and opposite forces

Electric charge The strength of the force is given by Coulomb’s Law The force is proportional to the magnitude of the charges 𝑞 1 , 𝑞 2 The force is inversely proportional to the square of the separation 𝑟 The force acts along the line joining the charges : by symmetry, no other direction could be singled out The force strength is governed by the permittivity of free space 𝜀 0 where 1 4𝜋 𝜀 0 =9× 10 9 𝑁 𝑚 2 𝐶 −2 𝑭 = 𝒒 𝟏 𝒒 𝟐 𝟒𝝅 𝜺 𝟎 𝒓 𝟐 𝒓 𝐹 𝑞 1 𝑟 𝑞 2 𝐹

Vectors A vector is a quantity – such as a force – which has both a magnitude and a direction It can be indicated by 𝐹 or F or F (helpful!) A vector can be specified by its components along co- ordinate axes, such as 𝐹 =( 𝐹 𝑥 , 𝐹 𝑦 , 𝐹 𝑧 ) The magnitude of a vector is 𝐹 = 𝐹 𝑥 2 + 𝐹 𝑦 2 + 𝐹 𝑧 2 A unit vector, indicated by 𝐹 , has magnitude =1 You may also see 𝐹 = 𝐹 𝑥 𝑖 + 𝐹 𝑦 𝑗 + 𝐹 𝑧 𝑘 , in terms of unit vectors along co-ordinate axes

Vectors Vectors may be added by summing their components The dot product of two vectors, 𝑎 . 𝑏 , is the projection of one vector along the other, 𝑎 𝑏 cos 𝜃 (𝜃= angle between). It can also be evaluated as 𝑎 . 𝑏 = 𝑎 𝑥 𝑏 𝑥 + 𝑎 𝑦 𝑏 𝑦 + 𝑎 𝑧 𝑏 𝑧 The cross product of two vectors, 𝑎 × 𝑏 , is a vector perpendicular to both with magnitude 𝑎 𝑏 sin 𝜃

Electric charge The total force from multiple charges is given by the principle of superposition 𝐹 13 𝐹 1 = 𝐹 12 + 𝐹 13 = 𝑞 1 𝑞 2 4𝜋 𝜀 0 𝑟 12 2 𝑟 12 + 𝑞 1 𝑞 3 4𝜋 𝜀 0 𝑟 13 2 𝑟 13 𝐹 12 𝑞 1 𝑟 12 𝑟 13 𝑞 2 𝑞 3

Electric field How is a force communicated between charges? A useful model is the electric field We interpret that electric charges set up an electric field 𝐸 in the region of space around them Then, a “test charge” 𝒒, placed in the electric field, will feel a force 𝑭 =𝒒 𝑬 The electric field is a vector “force field” representing the size/direction of the force per unit charge

Electric field The electric field can be represented by field lines: Electric field lines start on positive charges and end on negative charges. Their direction shows the force acting on a test charge; their spacing indicates the force strength. They can never cross.

Electric field A more complicated example of an electric field: There are no electric field lines here, because a test charge experiences zero net force

Electric field The electric field around a point charge +𝑸 follows simply from Coulomb’s Law Place a test charge +𝑞 at distance 𝑟 from +𝑄 Force 𝐹 = 𝑄𝑞 4𝜋 𝜀 0 𝑟 2 𝑟 Electric field 𝑬 = 𝑭 𝒒 = 𝑸 𝟒𝝅 𝜺 𝟎 𝒓 𝟐 𝒓 𝑞 𝑟

Introducing Gauss’s Law For general distributions of charges, it would be a nightmare to work out the electric field through the superposition principle 𝐸 = 𝑖 𝑞 𝑖 4𝜋 𝜀 0 𝑟 𝑖 2 𝑟 𝑖 Luckily we can use a more powerful method, Gauss’s Law. This is equivalent to Coulomb’s Law, but more convenient Gauss’s Law says that for any closed surface 𝑺, the flux of the electric field 𝑬 through 𝑺 is equal to the total charge enclosed by 𝑺, divided by 𝜺 𝟎 In mathematical terms: 𝐸 . 𝑑 𝐴 = 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 0

Surface integrals We are familiar with the concept of integration as summing up the “area under a curve” – written as 𝑎 𝑏 𝑓 𝑥 𝑑𝑥

Surface integrals This can be generalized to more dimensions! A 2D integral sums up the volume under a surface, 𝑓 𝑥,𝑦 𝑑𝑥 𝑑𝑦

Surface integrals The flux over a surface sums up the projection (dot product) of a vector field across the surface First, we break the surface into area elements 𝑑 𝐴 (these are vectors, in the direction normal to the surface) 𝑑 𝐴 surface area element 𝑑𝐴

Surface integrals For each area element, we evaluate the dot product with the vector field, 𝐸 .𝑑 𝐴 =𝐸 𝑑𝐴 cos 𝜃 (𝜃= angle between) We sum this up over the surface to obtain the “flux” 𝐸 .𝑑 𝐴 Important : we will only ever consider cases where 𝑬 is perpendicular to the surface, so 𝑬 .𝒅 𝑨 =𝑬×𝑨𝒓𝒆𝒂 𝑑 𝐴 𝑑 𝐴 𝐸 This case has zero flux! 𝐸

Introducing Gauss’s Law 𝐸 . 𝑑 𝐴 = 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 0 Breaking it down … 𝑆 is any closed surface (i.e., with no edges/gaps) Consider a small area element ∆ 𝐴 of 𝑆, where the vector is normal to 𝑆 The flux of 𝐸 through the area element is equal to 𝐸 .∆ 𝐴 =𝐸 ∆𝐴 cos 𝜃 The surface integral 𝐸 .𝑑 𝐴 means the total flux of 𝐸 through the surface Gauss’s Law says that this is equal to the total charge enclosed divided by 𝜀 0 𝑆

Introducing Gauss’s Law 𝐸 . 𝑑 𝐴 = 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 0 Why is it the same as Coulomb’s Law? Let’s derive Gauss’s Law … From the definition of solid angle we know that ∆Ω= ∆𝐴 cos 𝜃 𝑟 2 Coulomb’s law : 𝐸= 𝑞 4𝜋 𝜀 0 𝑟 2 Flux through ∆ 𝐴 is 𝐸 .∆ 𝐴 = 𝐸 ∆𝐴 cos 𝜃 = 𝑞 ∆𝐴 cos 𝜃 4𝜋 𝜀 0 𝑟 2 = 𝑞 ∆Ω 4𝜋 𝜀 0 Integrating over the whole surface: 𝐸 .𝑑 𝐴 = 𝑞 4𝜋 𝜀 0 𝑑Ω = 𝑞 𝜀 0 This holds true for any surface 𝑆 𝑆

Introducing Gauss’s Law 𝐸 . 𝑑 𝐴 = 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 0 Example: point charge +𝑄 By spherical symmetry, the electric field 𝐸 is radial Choose a spherical surface 𝑆 of radius 𝑟 centred on the charge 𝐸 has the same magnitude across the surface and cuts it at right angles, so 𝐸 .𝑑 𝐴 =𝐸×𝐴 Gauss’s Law then simplifies to: 𝐸×4𝜋 𝑟 2 = 𝑄 𝜀 0 Re-arranging: 𝑆 𝑟 𝐸= 𝑄 4𝜋 𝜀 0 𝑟 2

Summary Electric charge is a fundamental property of nature Electric charges 𝑞 1 , 𝑞 2 at a distance 𝑟 attract/repel with a Coulomb force 𝐹 = 𝑞 1 𝑞 2 4𝜋 𝜀 0 𝑟 2 𝑟 Electric charges set up an electric field 𝐸 in the region of space around them; a test charge 𝑞 placed in the field feels a force 𝐹 =𝑞 𝐸 A powerful method for deriving the electric field is Gauss’s Law for a closed surface, 𝐸 .𝑑 𝐴 = 𝑄 𝑒𝑛𝑐𝑙𝑜𝑠𝑒𝑑 𝜀 0