Gauss’s Law and Electric Potential

Gauss’s Law: Gauss’ law relates the electric fields at points on a (closed) Gaussian surface to the net charge enclosed by that surface. Gaussian surface, can have ANY shape, but the shape that minimizes our calculations of the electric field is one that mimics the symmetry of the charge distribution. The electric flux through a Gaussian surface is proportional to the net number of electric field lines passing through that surface.

E0 a What is the Flux of the field through the butterfly net ? a. E0pa2 b. 0 c. -E0pa2 d. Not enough Info

A) the sphere is replaced by a cube of the same volume A charged point particle is placed at the center of a spherical Gaussian surface. The electric flux Φ is changed if: A) the sphere is replaced by a cube of the same volume B) the sphere is replaced by a cube of one-tenth the volume C) the point charge is moved off center (but still inside the original sphere) D) the point charge is moved to just outside the sphere E) a second point charge is placed just outside the sphere

Electric Potential – Start by thinking about Potential Energy!!

Calculating the Potential from the Field: Since the electrostatic force is conservative, all paths yield the same result. If we set potential Vi =0, then

One Example – a Line of Charge: If l is the charge per unit length, then the charge on length dx is:

Calculating the Field from the Potential: Suppose that a positive test charge q0 moves through a displacement from one equipotential surface to the adjacent surface. The work the electric field does on the test charge during the move is W=-q0 dV. But we already said that the work is a scalar product of : Therefore, That is, Since E cos q is the component of E in the direction of ds, Therefore, the component of E in any direction is the negative of the rate at which the electric potential changes with distance in that direction.

Electrostatic Potential V is Scalar Relative quantity Units: J/C = Volt And it increases towards a + charge decreases towards a – charge

Equipotential Surfaces

Van de Graaff Generator Introduce pot and discuss large sphere vs small one – connect them …

Electrostatic Smoke Precipitators Strong E-fields are used to charge smoke and dust particles, which are then attracted to the walls and fall as ash precipitate. Demo (using van de Graff to bias) Physics 117 – Fall 2012

Ponderable – the Smokestack For this next summer you have an internship with LA Department of Water and Power. They have asked you to help design the air cleaners that will be used on a new coal burning power plant. Fly ash, which is very light (typically 1 * 10-4g) and small in diameter (typically 1mm), exits the boiler along with the hot gases. It is this fly ash with which you are concerned. Current plants are using electrostatic precipitators to remove the fly ash. They wish to continue using this technology as it removes over 99% of these particles, but you must redesign the system to match the specifications of the new plant. An ash precipitator removes solid emissions from smokestacks by electrostatic means. As the gas, laden with fly ash, leaves the boiler it is sent through a system that first gives the particles a net negative charge of 10-10C/particle. The gas then proceeds to a vertical stack. There the small particles move upward with the gases at 1m/s. In the stack, two oppositely charged parallel, vertical plates are placed so that the smoke passes through them. The plates are 15 meters tall and there is 1 meter between the plates. The horizontal electric field between the plates detects the ash and traps it on one of the plates. You want to know how powerful the electric field needs to be such that most of the particles deflect into the plates and never make it out of the stack. If the plates are square, what charge must each plate have on it to produce that electric field? What voltage is required on a plate to produce this electric field?

ConcepTest 20.1a Electric Potential Energy I A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which feels the larger electric force? 1) proton 2) electron 3) both feel the same force 4) neither – there is no force 5) they feel the same magnitude force but opposite direction Since F = qE and the proton and electron have the same charge in magnitude, they both experience the same force. However, the forces point in opposite directions because the proton and electron are oppositely charged. electron proton + -

ConcepTest 20.1b Electric Potential Energy II A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. Which has the larger acceleration? 1) proton 2) electron 3) both feel the same acceleration 4) neither – there is no acceleration 5) they feel the same magnitude acceleration but opposite direction electron proton + - Since F = ma and the electron is much less massive than the proton, then the electron experiences the larger acceleration.

ConcepTest 20.1c Electric Potential Energy III A proton and an electron are in a constant electric field created by oppositely charged plates. You release the proton from the positive side and the electron from the negative side. When it strikes the opposite plate, which one has more KE? 1) proton 2) electron 3) both acquire the same KE 4) neither – there is no change of KE 5) they both acquire the same KE but with opposite signs electron proton + - Since PE = qV and the proton and electron have the same charge in magnitude, they both have the same electric potential energy initially. Because energy is conserved, they both must have the same kinetic energy after they reach the opposite plate.

ConcepTest 20.4 Hollywood Square 1) E = 0 V = 0 2) E = 0 V  0 3) E  0 V  0 4) E  0 V = 0 5) E = V regardless of the value Four point charges are arranged at the corners of a square. Find the electric field E and the potential V at the center of the square. The potential is zero: the scalar contributions from the two positive charges cancel the two minus charges. However, the contributions from the electric field add up as vectors, and they do not cancel (so it is non-zero). -Q +Q Follow-up: What is the direction of the electric field at the center?

ConcepTest 20.5a Equipotential Surfaces I 1 3 2 4 +Q –Q At which point does V = 0? 5) all of them All of the points are equidistant from both charges. Since the charges are equal and opposite, their contributions to the potential cancel out everywhere along the mid-plane between the charges.

ConcepTest 20.6 Equipotential of Point Charge 1) A and C 2) B and E 3) B and D 4) C and E 5) no pair Which two points have the same potential? Since the potential of a point charge is: only points that are at the same distance from charge Q are at the same potential. This is true for points C and E. They lie on an Equipotential Surface. A C B D E Q

ConcepTest 20.7b Work and Electric Potential II 5) all require the same amount of work Which requires zero work, to move a positive charge from P to points 1, 2, 3 or 4 ? All points are the same distance from P. For path #3, you are moving in a direction perpendicular to the field lines. This means you are moving along an equipotential, which requires no work (by definition). P 1 2 3 4 Follow-up: Which path requires the least work?

24.4.5. Consider the equipotential lines shown in the box. The labeled cases indicate electric field line drawings. Which of these cases best matches the equipotential lines shown? a) 1 b) 2 c) 3 d) 4 e) None of these cases match the equipotential lines shown.

24.9.1. Consider two conducting spheres with one having a larger radius than the other. Both spheres carry the same amount of excess charge. Which one of the following statements concerning the potential energy of the two spheres is true? a) The potential energy of the larger sphere is greater than that of the smaller sphere. b) The potential energy of the larger sphere is the same as that of the smaller sphere. c) The potential energy of the larger sphere is less than that of the smaller sphere.