NYJC EJC Electric and Magnetic Fields Notes

9814 H3 Physics (2026) 1 ELECTRIC AND MAGNETIC FIELDS Content • Electric fields in a conductor • Gauss’s law for electric and magnetic fields • Ampère’s law for magnetic fields • Electric and magnetic dipoles Learning Outcomes Candidates should be able to: (a) show an understanding that ideal conductors form an equipotential volume and that the electric field within an ideal conductor is zero. (b) show an understanding that electric charge accumulates on the surfaces of a conductor and that the electric field at the surface of a conductor is normal to the surface. (c) recall and apply Gauss’s law for electric and magnetic fields (knowledge of the differential form of Gauss’s law is not required) and (i) solve problems involving symmetric charge distributions by relating the electric flux (in a vacuum) through a closed surface with the charge enclosed by that surface. (ii) show an understanding of the non -existence of “magnetic charge” expressed by Gauss’s law for magnetism. (d) recall and apply Ampère’s law relating the line integral of the magnetic field (in a vacuum) around a closed loop with the electric current enclosed by the loop to solve problems involving symmetric field configurations (knowledge of the differential form of Ampère’s law is not required). [Note further that candidates are not required to know Maxwell’s generalisation of Ampère’s law including the term related to the rate of change of electric flux, nor the Biot-Savart law.] (e) define the magnitude of the electric dipole moment as the product of the charge and the separation. (f) show an understanding of and use the torque on an electric dipole and the potential energy of an electric dipole to solve related problems. (g) define the magnitude of the magnetic dipole moment for a current loop as the product of the current and the area of the loop. (h) show an understanding of and use the torque on a magnetic dipole and the potential energy of a magnetic dipole to solve related problems. (i) appreciate that while electric and magnetic dipoles behave analogously, the theoretical framework at this level of study does not admit the possibility of magnetic monopoles.

9814 H3 Physics (2026) 2 show an understanding that ideal conductors form an equipotential volume and that the electric field within an ideal conductor is zero. Electric fields in a conductor When an electrical conductor is charged, all the charges (are able to move freely) will immediately move about inside the conductor due to mutual repulsion. The charges will very quickly redistribute themselves and stop moving reaching electrostatic equilibrium. This happens when the deposited charges distribute themselves on the surface of the conductor in such a way that: (a) the electric field inside the conductor is zero (b) the electric field at every point of the surface of the conductor is perpendicular to the surface. i.e. The electric potential inside the conductor is constant and is equal to the electric potential at the surface  the entire conductor is an equipotential volume. Conducting Sphere Since charges on a spherical conductor will be uniformly distributed on its surface, the charge distribution will also be spherically symmetric. Hence E and V outside the sphere are given by the same expressions as that of a point charge. Outside of the sphere: 2 04= QE r , 04= QV r Inside of the sphere: Since =− dVE dr , the electric field within the conductor must be zero as there is no potential difference within (or on the surface of) the conductor  potential V within and on the surface of the conductor must be uniform. V r R R R E r +Q

9814 H3 Physics (2026) 3 show an understanding that electric charge accumulates on the surfaces of a conductor and that the electric field at the surface of a conductor is normal to the surface. Note: 1. For a conductor in electrostatic equilibrium, all charges reside on the surface and E field is zero everywhere inside the conductor regardless of the shape and size of the conductor. 2. The E field is zero everywhere inside the conductor (regardless whether solid or hollow). 3. The E field outside a charged spherical conductor is the same as if all the charge is located at the centre of the sphere (i.e. point charge). 4. Since there is no E field within a conductor, and the entire conductor is at the same electric potential, no work is done to move a charge through it. 5. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature of the surface is smallest i.e. charges are closer together at sharp points. The E field looks as if due to a point charge, with the field lines spreading out radially.  the field is stronger near the sharp tip than on the flat part of the conductor. 6. The E field at a point just outside a charged conductor is perpendicular to the surface of the conductor and has a magnitude 0   ( = surface charge density at that point) . Continuous Charge Distribution When the distances between charges in a group of charges is much smaller than the distance between the group to a point where the electric field strength or electric potential is to be calculated, the system of charges can be modelled as a continuous charge distribution. The electric field strength or electric potential due to a continuous charge distribution must be evaluated for each infinitesimally small charge element dq. The electric field strength (vector) and electric potential (scalar) at a distance r due to a charge element dq is 2 0 ˆ 4 dqd r=Er and 04 dqdV r= respectively, where ˆr is a unit vector directed towards that point. The total electric field strength and electric potential due to the continuous charge distribution is therefore 2 0 1 ˆ 4 dq r= Er and 0 1 4 dqV r=  When we determine the electric field strength and electric potential due to various charge distributions, it is convenient to use the concept of charge density defined in the following ways: If a charge Q is uniformly distributed dq r dE P Q

9814 H3 Physics (2026) 4 • along a line of length L, the linear charge density  is QL = • on a surface of area A, the surface charge density  is QA = • within a volume V, the volume charge density  is QV = Example 1 A thin rod of length l and uniform charge per unit length  lies along the x axis. (a) Show that the electric field at P, a distance y from the rod along its perpendicular bisector and is given by 0 0 sin 2  =E y . (b) Using your part above, show that the field of a rod of infinite length is 02  =E y . Suggestion: First calculate the field at P due to an element of length dx, which has a charge of  dx. Then change variables from x to , using the relationships tan=xy and 2sec d=dx y , and integrate over . Solution: (a) The electric field at point P due to each element of length dx is 22 04 ( )= + dqdE xy