Electric Flux
Electric flux is the Chapter 24 product of the magnitude of the electric field and the Gauss’s Law surface area, A, perpendicular to the field
ΦE = EA
Defining Electric Flux Electric Flux, General Area
EFM06AN1 The electric flux is proportional to the number of electric field lines penetrating some surface
The field lines may make some angle θ with the perpendicular to the surface
Then ΦE = EA cos θ
1 Electric Flux, Interpreting the Equation Electric Flux, General
The flux is a maximum when the In the more general surface is perpendicular to the field case, look at a small The flux is zero when the surface is area element
parallel to the field E E iAi cos i E i A i
If the field varies over the surface, then In general, this Φ = EA cos θ is valid for only a small becomes lim E A EA d element of the area E i i Ai 0 surface
Electric Flux, final Electric Flux, Closed Surface
The surface integral means the integral Assume a closed must be evaluated over the surface in surface question The vectors ΔAi point in different In general, the value of the flux will directions depend both on the field pattern and on At each point, they the surface are perpendicular to . 2 the surface The units of electric flux will be N m /C By convention, they 2 i.e. [N/C] x [m ] point outward
2 Quick Quiz 24.1
Flux Through Closed Surface Suppose the radius of a sphere, with a charge at its centre, is halved. What happens to the flux through the sphere and the The net flux through the surface is magnitude of the electric field at its surface? proportional to the net number of lines leaving (a) The flux and field both increase. the surface (b) The flux and field both decrease. This is the number of lines leaving the surface minus the number entering the surface (c) The flux increases and the field decreases.
If En is the component of E perpendicular to (d) The flux decreases and the field increases. the surface, then (e) The flux remains the same and the field increases. (f) The flux decreases and the field remains the same. E E dA E n dA
Quick Quiz 24.1 Quick Quiz 24.2
Answer: (e). The same number of field lines pass through a In a charge-free region of space, a closed container is placed sphere of any size. Because points on the surface of the in an electric field. A requirement for the total electric flux sphere are closer to the charge, the field is stronger. through the surface of the container to be zero is that (a) the field must be uniform (b) the container must be symmetric (c) the container must be oriented in a certain way (d) The requirement does not exist – the total electric flux is zero no matter what.
3 Quick Quiz 24.2
Answer: (d). All field lines that enter the container also leave Gauss’s Law, Introduction the container so that the total flux is zero, regardless of the nature of the field or the container. Gauss’s law is an expression of the general relationship between the net electric flux through a closed surface and the charge enclosed by the surface
The closed surface is often called a Gaussian surface
Gauss’s law is of fundamental importance in the study of electric fields
Gauss’s Law – General Gauss’s Law – General, cont.
A positive point The field lines are directed radially outward charge, q, is located and are perpendicular to the surface at at the centre of a sphere of radius r every point The magnitude of E E dA E dA the electric field This will be the net flux through the everywhere on the gaussian surface, the sphere of radius r surface of the 2 2 sphere is We know E = keq/r and Asphere = 4πr , 2 E = keq / r q E E Asphere 4k eq 0
4 Gauss’s Law – General, notes Gauss’s Law – Final q The net flux through any closed surface Gauss’s law states in E E dA surrounding a point charge, q, is given by q/εo 0 and is independent of the shape of that surface. The net electric flux through a closed surface that qin is the net charge inside the surface surrounds no charge is zero. E represents the electric field at any point on Since the electric field due to many charges is the surface the vector sum of the electric fields produced by E is the total electric field and may have contributions the individual charges, the flux through any from charges both inside and outside of the surface closed surface can be expressed as Although Gauss’s law can, in theory, be solved to find E for any charge configuration, in E E dA E1 E 2 dA practice it is limited to symmetric situations
Finding an electric field using Quick Quiz 24.3
Gauss’s law If the net flux through a gaussian surface is zero, the following four statements could be true. Which of the EFA06AN1 statements must be true? (a) There are no charges inside the surface. (b) The net charge inside the surface is zero. (c) The electric field is zero everywhere on the surface. (d) The number of electric field lines entering the surface equals the number leaving the surface.
5 Quick Quiz 24.3 Quick Quiz 24.4
Answer: (b) and (d). Consider the charge distribution shown in the figure. The charges contributing to the total electric flux through surface Statement (a) is not necessarily true because an equal S’ are number of positive and negative charges could be present inside the surface. (a) q1 only
Statement (c) is not necessarily true - the number of field (b) q4 only lines entering the surface will be equal to the number (c) q and q leaving the surface, but this does not mean the field is zero 2 3 on the surface. (d) all four charges (e) none of the charges
Quick Quiz 24.4 Quick Quiz 24.5
Answer: (c). The charges q1 and q4 are outside the surface Again consider the charge distribution shown in this figure. and contribute zero net flux through S’. The charges contributing to the total electric field at a chosen point on the surface S’ are
(a) q1 only
(b) q4 only
(c) q2 and q3 (d) all four charges (e) none of the charges
6 Quick Quiz 24.5 Conditions for a Gaussian
Answer: (d). We don't need the surfaces to realize that any Surface given point in space will experience an electric field due to all local source charges. Choose a surface over which the integral can be simplified and the electric field determined. Try to choose a surface that satisfies one or more of these conditions:
The value of the electric field can be argued from symmetry to be constant over the surface . The dot product of E dA can be expressed as a simple algebraic product EdA because E and dA are parallel
The dot product is 0 because E and dA are perpendicular
The field can be argued to be zero over the surface
Field Due to a Spherically Field Due to a Point Charge Symmetric Charge Distribution
Choose a sphere as the Select a sphere as the Gaussian surface Gaussian surface
E is parallel to dA at each For r >a point on the surface
q Q E dA E dA E dA E dA E E 0 0
2 2 Q Q Q E E dA E (4 r ) E 4r so E k 4 r 2 e r 2 q q 0 0 E 2 k e 2 Q is the total charge inside the surface 4 0 r r
7 Spherically Symmetric Spherically Symmetric, cont. Distribution, final
Select a sphere as the Gaussian Inside the sphere, E surface, r < a varies linearly with r
3 E → 0 as r → 0 qenclosed = (4/3πr ) Charge density The field outside the = Q / [4/3πa3] sphere is equivalent to that of a point qenc E E dA E dA 0 charge located at 4 the centre of the r 3 sphere qenc qenc 3 4k e r k e Qr E 2 k e 2 k e 2 3 4 0 r r r 3 a
Field Due to a Thin Spherical Field at a Distance from a Line Shell of Charge
Use spheres as the Gaussian surfaces Select a cylindrical
When r > a, the charge inside the surface is Q and charge distribution E = k Q / r2 The cylinder has a e radius of r and a length When r < a, the charge inside the surface is 0 and E = 0 of ℓ E is constant in magnitude and perpendicular to the surface at every point on the curved part of the surface
8 Field Due to a Line of Charge Field Due to a Line of Charge
The end view Use Gauss’s law to find the field confirms the field is perpendicular to the q encl l E E dA E dA EA curved surface 0 0 The field through the A 2rl, so E (2rl) l ends of the cylinder 0 is 0 since the field is parallel to these E 2k 2 r e r surfaces 0
Field Due to a Plane of Charge Field Due to a Plane of Charge
E must be perpendicular to the plane The total charge in the surface is σA and must have the same magnitude at all points equidistant from the plane Applying Gauss’s law
Choose a small cylinder whose axis is A 2 EA , so E perpendicular to the plane for the E 2 Gaussian surface 0 0
E is parallel to the curved surface, with Note, this does not depend on r no contribution to the surface area from this curved part of the cylinder Therefore, the field is uniform everywhere
The flux through each end of the cylinder is EA and so the total flux is 2EA
9 Properties of a Conductor in Electrostatic Equilibrium Electrostatic Equilibrium
1. The electric field is zero everywhere inside When there is no net motion of charge the conductor. within a conductor, the conductor is said 2. If an isolated conductor carries a charge, the to be in electrostatic equilibrium charge resides on its surface. 3. The electric field just outside a charged conductor is perpendicular to the surface and has a magnitude of σ/εo. 4. On an irregularly shaped conductor, the surface charge density is greatest at locations where the radius of curvature is the smallest.
Property 2: Charge Resides
Property 1: Einside = 0 on the Surface
Consider a conducting slab in Choose a Gaussian surface an external field E inside but close to the actual surface If the field inside the conductor were not zero, free electrons in The electric field inside is zero (property 1) the conductor would experience an electrical force There is no net flux through the Gaussian surface These electrons would Because the Gaussian accelerate surface can be as close to These electrons would not be the actual surface as in equilibrium desired, there can be no charge inside the surface Therefore, there cannot be a field inside the conductor Therefore, any charge must be on the surface.
10 Demo EA10: Gauss’s law; field Property 3: Field’s Magnitude inside insulator, not conductor and Direction
Excess charges resides on the outer surface of a Choose a cylinder as the conductor. A metal can is placed on top of a van Gaussian surface der Graaff generator and filled with styrofoam The field must be packing. When the generator is turned on, the perpendicular to the packing remains within the cup. One can show that surface, for, if there were the outside of the can has charge by drawing off a a parallel component to E, spark with the earth rod. The metal can and the charges would experience generator are all at one equipotential and the a force and accelerate charge is on the outside of the conductor. There is along the surface and it no field inside the can. The can is replaced by a would not be in polystyrene cup. In this case, the packing is ejected equilibrium. out of the cup. Because the cup is an insulator, a A field can exist within it – the cup makes little E EA , and E disturbance to the equipotential surfaces. 0 0
Property 4: Charge density highest at sharp points
The charge density is high where the radius of curvature is small End of Chapter
And low where the radius of curvature is large
The electric field is large near the convex points having small radii of curvature and reaches very high values at sharp points
11 Quick Quiz 24.6 Quick Quiz 24.6
Your little brother likes to rub his feet on the carpet and then Answer: (a). Charges added to the metal cylinder by your touch you to give you a shock. While you are trying to brother will reside on the outer surface of the conducting escape the shock treatment, you discover a hollow metal cylinder. If you are on the inside, these charges cannot cylinder in your basement, large enough to climb inside. In transfer to you from the inner surface. For this same reason, which of the following cases will you not be shocked? you are safe in a metal automobile during a lightning storm. (a) You climb inside the cylinder, making contact with the inner surface, and your charged brother touches the outer metal surface. (b) Your charged brother is inside touching the inner metal surface and you are outside, touching the outer metal surface. (c) Both of you are outside the cylinder, touching its outer metal surface but not touching each other directly.
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