Electric fields Electric charge is a fundamental characteristic of matter. Point charges (i.e. that are infinitesimally small with respect to the characteristic distance scales of interaction) will interact with each other according to Coulomb’s Law. This is an inverse-square law, and directly analogous to Newton’s law of Universal Gravitation..

Electric field produced by Particle Charge Mass /kg Mass / electron e 1.60221019 C a positive charge. (For a masses Charge is measured in negative charge, lines of coulombs, with the 27 force go inward) Proton + e 1.6726 10 1,836 fundamental unit being q the charge on the Q Neutron 0 1.6749 1027 1,839 F electron. + + F Electron - e 9.1094 1031 1 Charles-Augustin de Coulomb 1736-1806 Like charges will repel It is instructive to imagine the possible influence of a charge with a force on another, if the second charge could be at any location in space. 1 qQ This is This is the concept of an electric field E. At any point in space we can F  imagine a vector describing the magnitude and direction of the 4 r2 Coulomb’s 0 Law of force per unit charge, acting on a charge q. 12 1 Electrostatics r 0  8.8542 10 Fm FE q Permittivity of free space Using Newton’s Second Law, the charge would accelerate in the direction of F. The vector field E therefore tells us where a charge Compare the strength of gravity for two protons would move at a point in space. This is just like a magnetic field describing the direction that a compass needle would point at any at separation d with electrostatic repulsion Carl Friedrich Gauss point in space, but also how strong the force is at any point. 1 e2 1777-1855 FE  2 4 0 d The electric field sourced by a point Gm2 So how does a nucleus F  charge Q is given by: G d 2 stay together? Over scales of 10-15 m F e2 1 Q E  the Strong and Weak Er ˆ rˆ is a unit vector F 4 Gm2 nuclear forces counteract 2 G 0 40 r pointing in the radial electrostatic repulsion. direction away F 36 E 1.26 3 10 Neutrons as well as protons from the point charge Q FG contribute to these forces.

For two electrons: So the field lines due to the point Michael Faraday charge radiate outwards, if Q is positive. F 36 2 1791-1867 E 1.236  10  1,836 The field line divergence is indicative of FG So electric force is the decay in strength of E with radius r F E 4.165 1042 MUCH stronger at FG an atomic scale. The field lines indicate the direction the positive charge would move at this point in the electric field due to the charge in the centre of the diagram

G  6.6741 1011 m 3 kg -1 s -2 Universal Gravitational constant Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 1 If there is more than one point charge, the vector fields will superpose, i.e. add in a vector sense. A Dipole is an arrangement of two opposite charges, whereas a parallel plate capacitor can be modelled by two lines of opposite point charges, separated by a fixed distance. In all the examples below, a MATLAB computer program has added up the E fields due to the point charges for any x,y location. Notice the field strength between the capacitor plates appears to be constant. (,)xy ˆˆ 1 Qi xyx Xii  y  Y  E(,)xy  22  22i.e. charge Qi at fixed location  XYii,  Qi 4 0 i x X  y  Y  ii   x Xii  y  Y 

Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 2 Gauss’ Law Electric potential Rather than adding up point charges, we can use Gauss’s law to The work done W in bringing a charge from an infinite distance to a specific position in an electric field work out the electric field which passes through a surface S which is proportional to the electric potential V. This quantity can be used in Conservation of Energy calculations encloses a charge Q The electric potential is the electrical energy per unit charge, i.e. voltage. Q ESd FE q force on charge q For the electric potential at radius r from a point charge of strength Q S  0 in electric field E r W WF  d r V  d r   dr rˆ For more complex distributions of charges,  Q    x,, y z dxdydz q it is often easier to compute the electric potential since this is a scalar function. The qQr qQ Carl Friedrich Gauss Fq E  rˆ  W   dr resulting electric field can be found from the 1777-1855 22 gradient of the potential charge density 4400rr  Q  ESd  VVV   zˆ S QQ11 EEE       0 V  dr  V   x y z r 2  x  y  z dSr ˆ dA 4400rr  r E  V Q dA rsin d  d  V   r Er Er()ˆ 4 r 0 This is called the gradient (‘grad’) vector yˆ ESd4 r2 E A useful result is that the operator. In other words, we have to find  S line integral of electric fields the slope of the electric potential in each Q BBB VVV   E from position A to B is the of the x,y,z directions. ˆ 2 Ed l V d l xˆ y ˆ  zˆ x ˆ dx y ˆ dy zˆ dz x difference in the electric AAA    4 0r x  y  z qQ potential, regardless of the i.e. Gauss’s Law is Fr ˆ path taken. BBBVVV   consistent with 2 El d   dx  dy  dz   dV  V()() A  V B B 4 0r AAA   Eld  V()() A  V B x  y  z A Coulomb’s Law for a Fields that work in this way  point charge are called conservative.

In this situation, both charges in the dipole are positive, so the field lines diverge. Note there is a region between the charges where the electric field is very small. Indeed it tends to zero at the very centre

Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 3 Electric field strength above an infinite conducting plane Parallel plate capacitor

Consider an infinite conducting plane with charge per unit area . A parallel plate capacitor can be modelled The electric field at a perpendicular distance z from the plane must as two infinite charges planes separated only be in the z direction. Why? If we consider concentric rings of by distance d, separated by an insulating charge contributing to the electric field, by symmetry we can clearly dielectric, and rolled up like a Swiss cheese.

see that any field in the x,y direction will be opposed by an equal Noting there are two planes (with opposing and opposite signed field from a charge at the diametric opposite in the circle. but equal charges Q ) zˆ Q    A A real capacitor will have finite area A. Let us assume this is  E 2 much larger than the plate  2 0 separation d so the ‘infinite Concentric z Q plane’ argument has some ring of E  validity charge  A Charge element 0  r dQ rdrd i.e. a constant electric field between the plates We can obtain the same result using Gauss’ law: xˆ Q ESd S  0 EExy0 Q Assuming the electric field  2 rd dr is both constant and perpendicular E  cos EA  z 22 to the conducting plane surface r00  0 4 0 rz  Q r22 hcos z E   0 A  2 z rd dr Ez 3 r00 2 4 rz22 0   The electric potential is: We can define the Capacitance (units are 2 z rdr d Farads) as the ratio of charge Ez  3 V Edx r0 2 4 22 0 stored on the plates to 0 rz  Qd voltage between them  V   z 1 rdr 1  A Ez  0 Q CV 22 3    c  2 22 2 0 rz 0 22 rz  rz V Q E C   z 1 d V Ez 0   2 0 z  A Hence: C  0  E  So the electric field strength above an infinite d z conductive plane with charge per unit area  is 2 0 a constant

For an excellent description of Capacitors see the MIT Physics 9.02 Electricity & Magnetism course Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 4 Electric field of a dipole* Electric field of a line charge Note the plot on the right Electric field outside a charged sphere of charge Q is only an approximation. 1 qdr ˆ Er Er()ˆ Instead of a sphere there Vr() is a ring of positive charges. 4 r2 d is the vector separation dz 0 of opposing charges While these will result in a zero rrˆ field strength at the centre, this 3qqd rˆˆ r d of magnitude q E()r  3 is not the case for all positions 4 0r inside the ring. This shortcoming has been ‘overcome’ (!) by setting the field strength to be zero inside Charges on a conducting the sphere. line will distribute themselves

until there is no net For a realistic simulation of a electric field along the line. One charged sphere, charges would expects this steady state situation have to be uniformly distributed to happen rapidly. on the surface. This is a three

dimensional distribution. Due to Therefore the electric field the inverse square law, this is perpendicular to the line, and means the field, due to all the will only vary with radial distance charges, is zero everywhere from the line, assuming the q d q inside the conductor. line is both long and straight.

Let the charge per unit length be l

Gauss’s Law The charged sphere will exhibit general features of any Q hollow conductor. Unless it is polarized via the application ESd of an external electric field, the charges will tend to S Er Er()ˆ  0 distribute uniformly on the conductor surface. If this was ldz dz not the case, any inhomogeneities would cause charge to 2 rEdz  move. Since charge is highly mobile in a conductor, we  0 rrˆ may assume a steady state is rapidly attained. Capacitance of a charged sphere Consider a charged sphere of charge Q placed l E Gauss’s Law – outside the conductor inside a hollow conducting sphere of charge -Q 2 r 0 Q b ESd bbQQ1 S  V V  V  Edr  dr   If the line charge was actually a cylindrical conductor of radius 0 abaa44rr2   00a a, and this was placed inside a hollow cylindrical conductor of QQi.e. a spherical E4 r2   E  Q11  Q  b a  radius b such that b > a, the voltage between the outer and inner 4  r2 conductor looks just like V       conductors would be 00a point charge 44a b    ab  B 00 Eld  V  V A abThis means the Q4 ab Gauss’s Law – inside the conductor C  0 capacitance per V b a bbl unit length of a ESd 0 i.e. there is no electric field inside V V  V  Edr  dr coaxial cable is: S abaa a hollow conductor. This is why a 2 0r 2 l2  E4 r  0  E  0 conductive container can be used if ba l b C  0 as an effective shield of EM V ln  V b Ca 4 2 a ln  radiation. We call this a Faraday 0 0 a Cage.

*This shall be justified in another Eclecticon note Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 5 Conductive sphere in a uniform electrical field The electric potential a long way from the sphere A conductive sphere placed in a uniform electric field will must relate to the electric field by be polarized by it. In other words, charges will be separated on the surface of the sphere to align with the field. V E  0 y Note unless the sphere was initially charged,

the total charge must still sum to zero. Either way, although Vuniform   E00 y   E r cos the distribution of charge is modified by the external field, the total amount on the sphere must remain the same. The full potential is therefore, for ra To determine the electrical field outside the sphere (if we assume a hollow conductor, then the field 1p cos V E r cos must be exactly zero inside) we shall make a sensible 4 r2 0 guess that it comprises of something which looks like 0 a dipole, superimposed upon a uniform field. We can therefore evaluate the electric field. Given the form of V it is most convenient to yˆ  do this in plane polar coordinates* ˆ (,)xy r + + + + ˆ E  V + + θ + + ˆ r ErEEr ˆ  θ  Vp2 cos EE  cos  xˆ r 0 3 Note to aid in plotting this field rr4 0 Simulation of the electric field a around a conducting sphere Conducting sphere of 1Vp sin ˆ ˆˆ 22 r xsin y cos r x y 11 -1 - - EE    0 sin  3 radius a rr4  0 8.85 10 Fm - - - 0 ˆ Ey E ˆ - θ xˆˆcos y sin x 4 -1 0 - - 1 E  10 Vm   tan  0 External field Now there cannot be any tangential y Induced electric fields at the surface of the conductor. a  0.05m charge If there were, then the charges would move. Gauss’ Law states separation Q E  r  a  0 ESd S  Imagine a dipole corresponding to two opposing psin 0 Charge distribution on charges placed at the top and bottom of the sphere E sin   0 0 3  a conducting sphere 4 a 1 qdr ˆ 0 Er () r  a  polarized by a 3  0 Vdipole  2 uniform electric field p4 00 a E 4 0 r where  is the charge per unit area 1 2qa cos on the sphere surface. ˆ ˆ 3 d2a y  d  r  2 a cos  Vdipole  2 2a 4 r Hence: EE1 cos 0 r 3 0 Hence: r Let us assume the ‘sphere-dipole field’ ra 3 3 has a similar expression, where ‘polarization’ p a 2a EE1 sin EEr 13 0 cos  3 0  is to be found: r r E( r  a )  3 E cos 1p cos r 0 Vdipole  2 ra EEr  0 3 00E cos  4 0 r

VVVV  1  * V xˆˆ  y  rˆ  θˆ Physics topic handout – Electrical fields Dr Andrew French. www.eclecticon.info PAGE 6 x  y  r r 