a b c

Gauss’ Law … made easy To solve the above equation for E, you have to be able to CHOOSE A CLOSED SURFACE such that the integral is TRIVIAL. (1) Direction: surface must be chosen such that E is known to be either parallel or perpendicular to each piece of the surface; If then (2) Magnitude: surface must be chosen such that E has the same value at all points on the surface when E is perpendicular to the surface.

Choose either or And E constant over surface is just the area of the Gaussian surface over which we are integrating. Gauss’ Law This equation can now be solved for E (at the surface) if we know q enclosed (or for q enclosed if we know E ). Gauss’ Law

Geometry and Surface Integrals If E is constant over a surface, and normal to it everywhere, we can take E outside the integral, leaving only a surface area z L R R you may use different E ’s for different surfaces of your “object” a b c x y z

Gauss  Coulomb We now illustrate this for the field of a point charge and prove that Gauss’ Law implies Coulomb’s Law. Symmetry  E -field of point charge is radially and spherically symmetric Draw a sphere of radius R centered on the charge. E +Q+Q R Why? E normal to every point on the surface E has same value at every point on the surface can take E outside of the integral!  

Gauss  Coulomb Therefore, –Gauss’ Law –We are free to choose the surface in such problems… we call this a “Gaussian” surface E +Q+Q R

Uniform charged sphere Outside sphere: ( r > a ) –We have spherical symmetry centered on the center of the sphere of charge –Therefore, choose Gaussian surface = hollow sphere of radius r What is the magnitude of the electric field due to a solid sphere of radius a with uniform charge density  (C/m 3 )? a  r  Gauss’ Law same as point charge!

Uniform charged sphere Outside sphere: ( r > a ) Inside sphere: ( r < a ) –We still have spherical symmetry centered on the center of the sphere of charge. –Therefore, choose Gaussian surface = sphere of radius r Gauss’ Law But, Thus: ra E a  r

Gauss’ Law and Conductors We know that E=0 inside a conductor (otherwise the charges would move until net field=0). But since . Charges on a conductor only reside on the surface(s)! Conducting sphere

Gauss’ Law and Conductors The electric field immediately outside a conductor must be perpendicular to the conductor surface. –Otherwise the charges would move along the surface until the field was perpendicular everywhere. Applying Gauss’ law to a small Gaussian cylinder perpendicular to the surface The field just outside a conductor is perpendicular to the surface and proportional to the surface charge density 

AB A blue sphere A is contained within a red spherical shell B. There is a charge Q A on the blue sphere and charge Q B on the red spherical shell. The electric field in the region between the spheres is completely independent of Q B the charge on the red spherical shell. True False Question 1

1.a 2.b 10 0 of 5

AB A blue sphere A is contained within a red spherical shell B. There is a charge Q A on the blue sphere and charge Q B on the red spherical shell. The electric field in the region between the spheres is completely independent of Q B the charge on the red spherical shell. True False Question 1

Infinite Line of Charge, charge/length= Symmetry  E -field must be  to line and can only depend on distance from line Therefore, CHOOSE Gaussian surface to be a cylinder of radius r and length h aligned with the x -axis. Apply Gauss’ Law: On the ends, On the barrel,  NOTE: we have obtained here the same result as we did previously using Coulomb’s Law. The symmetry makes today’s derivation easier x y ErEr h ErEr

Charge density on a conducting cylinder A line charge (C/m) is placed along the axis of an uncharged conducting cylinder of inner radius r i = a, and outer radius r o = b as shown. –What is the value of the charge density  o (C/m 2 ) on the outer surface of the cylinder? a b    View end on: Draw Gaussian tube contained within the conducting cylinder The field within the conducting cylinder is zero A charge equal and opposite to the line charge is induced on the inner conductor surface that cancels the line charge b oo

Charge density on a conducting cylinder a b    Now draw Gaussian tube which surrounds the outer edge The tube still contains the line charge and the field is the same as calculated before oo b r The charge inside the Gaussian tube also now =( charge on the outer surface =  0 2  bL ) + (charge on inner surface + line charge =0 ) Therefore by Gauss’ Law A charge equal to the line charge is induced on the outer surface of the cylinder

Question 2 Consider the following two topologies: A) A solid non-conducting sphere carries a total charge Q = -3  C distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. Compare the electric field at point X in cases A and B: (a) E A < E B (b) E A = E B (c) E A > E B E 22 11 -| Q| B) Same as (A) but conducting shell removed

Question 2 1.a 2.b 3.c 10 0 of 5

Question 2 Consider the following two topologies: A)A solid non-conducting sphere carries a total charge Q = -3  C distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. E 22 11 -| Q| Compare the electric field at point X in cases A and B: (a) E A < E B (b) E A = E B (c) E A > E B Select a sphere passing through the point X as the Gaussian surface. It encloses charge -|Q|, whether or not the uncharged shell is present. (The field at point X is determined only by the objects with NET CHARGE.) B) Same as (A) but conducting shell removed

Question 3 Consider the following two topologies: A) A solid non-conducting sphere carries a total charge Q = -3  C distributed evenly throughout. It is surrounded by an uncharged conducting spherical shell. E 22 11 -| Q| What is the surface charge density  1 on the inner surface of the conducting shell in case A? (a)  1 <  (b)  1 =  (c)  1 >  B) Same as (A) but conducting shell removed

Question 3 1.a 2.b 3.c 10 0 of 5

Consider the following two topologies: A solid non-conducting sphere carries a total charge Q = -3  C and is surrounded by an uncharged conducting spherical shell. B) Same as (A) but conducting shell removed What is the surface charge density  1 on the inner surface of the conducting shell in case A? (a)  1 <  (b)  1 =  (c)  1 >  E 22 11 Question 3 Inside the conductor, we know the field E = 0 Select a Gaussian surface inside the conductor Since E = 0 on this surface, the total enclosed charge must be 0 Therefore,  1 must be positive, to cancel the charge -|Q| By the way, to calculate the actual value:  1 = -Q / (4  r 1 2 ) -| Q|

Gauss’ Law proves that electric fields vanish in conductor –extra charges reside on surface Chapter 23 of Fishbane Try Chapter 23 problems 25, 29, 33, 47, 51, 56 Summary Gauss’ Law: Electric field flux through a closed surface is proportional to the net charge enclosed –Gauss’ Law is exact and always true…. Gauss’ Law makes solving for E -field easy when the symmetry is sufficient –spherical, cylindrical, planar