Gauss’ Law Field Lines Electric Flux Recall that we defined the electric field to be the force per unit charge at a particular point: P For a source point charge Q and a test charge q at point P q Q at P If Q is positive, then the field is directed radially away from the charge. + Note: direction of arrows Note: spacing of lines Note: straight lines If Q is negative, then the field is directed radially towards the charge. - Negative Q implies anti-parallel to Note: direction of arrows Note: spacing of lines Note: straight lines + + Field lines were introduced by Michael Faraday to help visualize the direction and magnitude of the electric field. The direction of the field at any point is given by the direction of the field line, while the magnitude of the field is given qualitatively by the density of field lines. In the above diagrams, the simplest examples are given where the field is spherically symmetric. The direction of the field is apparent in the figures. At a point charge, field lines converge so that their density is large - the density scales in proportion to the inverse of the distance squared, as does the field. As is apparent in the diagrams, field lines start on positive charges and end on negative charges. This is all convention, but it nonetheless useful to remember. - + This figure portrays several useful concepts. For example, near the point charges (that is, at a distance that is small compared to their separation), the field becomes spherically symmetric. This makes sense - near a charge, the field from that one charge certainly should dominate the net electric field since it is so large. Along a line (more accurately, a plane) bisecting the line joining the charges, we see that the field is directed along the -x direction as shown. + + In this case, we see the zero-field region precisely between the two charges, and we also see a fairly rapid convergence on a spherically symmetric distribution of field lines. Example (Question) The figure shows the electric field lines for a system of two point charges. (a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak? Example (Solution) The figure shows the electric field lines for a system of two point charges. (a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak? a) There are 32 lines coming from the charge on the left, while there are 8 converging on that on the right. Thus, the one on the left is 4 times larger than the one on the right. Example (Solution) The figure shows the electric field lines for a system of two point charges. (a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak? • The one on the left is positive; field lines leave it. The one on the right is negative; field lines end on it. Example (Solution) The figure shows the electric field lines for a system of two point charges. (a) What are the relative magnitudes of the charges? (b) What are the signs of the charges? (c) In what regions of space is the electric field strong? weak? • The field is strong near both charges. It is strongest on a line connecting the charges. Few field lines are drawn there, but this is for clarity. The field is weakest to the right of the right hand charge. END How would you measure 'the density of electric field lines' in a vicinity of space? First think only of a discrete set of electric field lines. One o b v i o u s a n s w e r t o t h e question is that you would count the number of lines passing through an imaginary geometrical (not real!) surface. The first two are obvious, and the Uniform electric field - the following diagram lines have constant density will indicate the and are all parallel last: The maximum number o f f i e l d l i n e s i s intercepted when the Surface is inserted in three surface is perpendicular different orientations and to the field, while no field lines will pass intersects the electric field through it when the lines. surface is parallel to the field. N u m b e r o f N u m b e r o f Number of field f i e l d l i n e s f i e l d l i n e s l i n e s p a s s i n g passing through passing through through surface is surface is large surface is zero between zero and the maximum. Depends on the orientation angle θ. Electric flux a function of surface orientation, … relative to the electric field In general, the number of field lines passing through an area A is directly proportional to cos(θ), where θ is the angle between the electric field vector and the normal to the surface vector. Use this symbol to represent electric flux through a surface Increasing surface area Increasing electric field The electric flux through the blue surface is the same in these two figures. Electric flux is proportional to the product of electric field and surface area A. The number of field lines passing through a geometrical surface of given area depends on three things: the field strength, the area, and the orientation of the surface. [E]-Electric field; Newton/Coulomb {N/C} [A]-Surface area; meter2 {m2} [θ]-Angle; degrees or radians {o, rad } 2 2 [ΦE]-Electric flux; Newton meter /Coulomb {Nm /C} Example (Question) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle. Area A Example (Solution) A uniform electric field of 2.1 kN/C passes through a rectangular area 22 cm by 28 cm. The field makes an angle of 30o with the normal to the area. Determine the electric flux through the rectangle. 30o 2.1kN/C 0.22 m X0.28 m = 0.062 m2 22 cm 28 cm END It is useful also to represent the area A by a vector. The length of the vector is given by the area (a scalar quantity), while the orientation is perpendicular to the area. With this definition, the flux can be defined as: recall A slightly different notation for electric flux: Unit vector perpendicular to surface of area A WHAT IF THE SURFACE IS CURVED AND/OR THE FIELD VARIES WITH POSITION? Element of surface area dA q Surface Consider the surface as made up of small elements over which the electric field is uniform. Flux for Non-Uniform Fields / Flux for Non-Uniform Surface You might have noticed that all these equations really only work for uniform electric fields. We can use them here provided we make them pertain to differential (small) area elements, and over a differential area the field is uniform. We then need to integrate (sum) to get the total flux through an extended surface in a non-uniform field. Flux for Non-Uniform Fields / Flux for Non-Uniform Surface The differential electric flux passing through a differential area is given by: Element of surface area dA As before, dA is a vector oriented perpendicular to the area, and the area itself is differential (i.e., it's infinitesimally small and it's shape doesn't usually matter). The total electric flux can be evaluated by integrating this differential flux over the surface. Element of surface area dA Flux for Non-Uniform Fields / Flux for Non-Uniform Surface Flux for one segment Surface Total flux through surface q THE INTEGRAL IS TAKEN OVER THE ENTIRE SURFACE. Divide surface into small elements dA THE BASIC DEFINITION OF ELECTRIC FLUX Example (Question) A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge. + At any point on the sphere the magnitude of the electric field is: + Point charge electric field at distance r from charge E = 6.75 X 105 N/C From symmetry the field is perpendicular to the spherical surface at every point. + Same direction on the surface The electric field has the same magnitude over the spherical surface. It is thus constant and may be taken outside the integral + The integral that remains is just the surface area of the sphere. The electric flux through the spherical surface is: + E = 6.75 X 105 N/C END Example (Question) A positive test charge of magnitude 3.0µC is surrounded by a sphere with radius 0.20 m centered on the charge. Find the electric flux through the sphere due to this charge. Consider the same example again, but this time we save the calculations for the + end. At any point on the sphere the magnitude of the electric field is: + Point charge electric field at distance r from charge The electric field has the same magnitude over the spherical surface. It is thus constant and may be taken outside the integral + The integral that remains is just the surface area of the sphere.