Previous Lectures: Introduced to Coulomb’s law Learnt the superposition principle Showed how to calculate the electric field resulting from a series of discrete charges Showed how to calculate the resultant force on a charge placed in an electric field

 To calculate E due to a continuous distribution of charges Learning Objective: Lecture 3: Electric fields Continuous? didn’t you say that charge is quantised? Yes, but water consists of individual H 2 O molecules, and it flows.

A Continuous Distribution of Charges Break the distribution into small pieces Treat each little piece as a point charge and add up the separate E-fields (as a vector of course!) This can get messy - typically results in a 3D integral, evaluated numerically by computer What do we do if the object is not a point particle?

Charge can be distributed along a line, over a surface, or through a volume. Use to represent charge/length over the element dl Use  to represent the charge/unit area over a surface area element dA Use  to represent the charge/unit volume in a volume element dV

E-field on the axis of a ring of charge (Tipler) Method: 1) Represent the charge distribution as a series of small elements ds each with a charge dQ (+ charge) dQ

1 2) Use symmetry. Consider two ring elements which are diagonal to each other 2 dE dQ  

x y No net E-field in the y direction dE 2 dE 1 dE Total P For one charge element

Note that for x >> a Coulomb’s Law Total E-field at P Charge distribution is uniform!  

The E-Field due to a large circular plane and uniform distribution of charges  1) Consider first an elemental ring

2) Calculate the net E-field at P due to the ring element:

3) Use the following trignometrical identities 

5) Consider the limiting case R   4) E-field due to the charged ring is Same result if x <<R

For an infinite plane sheet of charge the E-field produced is independent of the distance from the sheet This result is true close to the surface of any charge distribution

E-field at a point along a line perpendicular to the mid point of line of uniform charge. Student exercise: show that Note that for x >> a Coulomb’s Law dQ y No E y

Further exercise: Write = Q/2a Obtain an expression for E x as a   E field does not follow an inverse square law and is radially outward from the line

E-field inside a spherical shell of uniform charge Spherical Shell of Charge r1r1 r2r2 P a1a1 a2a2

Total E-Field at P is ZERO All parts of the surface can be paired off What if ?

Summary Illustrated how to calculate the electric field due to a continuous distribution of charge For an infinite plane sheet of charge the E-field produced is independent of the distance from the sheet E-field inside a spherical shell of uniform charge is zero

Classwork: Three charges, a, b, and c, each of 3  C, are located at x = 0, x = 0.5 m and x = 1.0 m, respectively. What is the force exerted by charges b and c on a? x abc

The principle of superposition of forces tells us that the force on charge a is the vector sum of the force caused by charge b and that caused by charge c – that is: F aresultant = F b + F c where F b & F c are respectively, the forces on charge a caused by charges b and c. Since all the charges are positive, the resultant force is in the –x direction.

Use Coulomb’s law:

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Spherically Symmetric Charge Distribution E outside a sphere or spherical shell carrying charge q is identical to that of a point charge q at the centre.