Electric Potential 2 q A C B r A B r path independence a a Rr VQ 4   r Q 4   R

Previously Work has to be done to move a charge in an electric field A charge has potential energy in the field Define potential energy (U) = charge (Q) x electric potential (V) Then A positive test charge will move toward lower potential Define zero potential at infinity Potential difference between two points

Question 1 Two test charges are brought separately to the vicinity of positive charge Q. –charge + q is brought to pt A, a distance r from Q. –charge +2 q is brought to pt B, a distance 2 r from Q. –Compare the potential at point A ( V A ) to that at B ( V B ): Q Q A q r B 2q2q 2r2r + + (a) V A < V B (b) V A = V B (c) V A > V B

THE POTENTIAL IS A FUNCTION OF THE SPACE !!!!! The Potential does not depend on the “test” charge at all. Question 1 Two test charges are brought separately to the vicinity of positive charge Q. –charge + q is brought to pt A, a distance r from Q. –charge +2 q is brought to pt B, a distance 2 r from Q. –Compare the potential at point A ( V A ) to that at B ( V B ): Q Q A q r B 2q2q 2r2r + + (a) V A < V B (b) V A = V B (c) V A > V B A positive “test charge” would move (“fall”) from point A towards point B. Therefore, V A > V B In fact, since point B is twice as far from the charge as point A, we calculate that V A = 2 V B !!

Potential Difference between two points We want to evaluate the potential difference from A to B in the field of a point charge q q A B r A B r E What path should we choose to evaluate the integral? If we choose straight line, the integral is difficult to evaluate. Magnitude of the field is different at each point along the line. Angle between E and path is different at each point along line.

Potential difference between two points We want to evaluate the potential difference from A to B in the field of a point charge q C E q A B r A B r From A to C, E is perpendicular to the path. i.e From C to B, E is parallel to the path. i.e Instead choose path A->C then C->B

Potential Difference between 2 points Evaluate potential difference from A to B along path ACB. q A C B r A B r E Evaluate the integral: by definition:

Independent of Path? How general is this result? Consider the approximation to the straight path from A->B (white arrow) = 2 arcs (radii = r 1 and r 2 ) plus the 3 connecting radial pieces. q A B r A B r C For the 2 arcs + 3 radials path: This is the same result as above!! The straight line path is better approximated by Increasing the number of arcs and radial pieces. q A B r1r1 r2r2

Potential from N charges The potential from a collection of N charges is just the algebraic sum of the potential due to each charge separately. x r1r1 r2r2 r3r3 q1q1 q3q3 q2q2 

+5 μC -3 μC 9) Two charges q 1 = + 5 μC, q 2 = -3μC are placed at equal distances from the point A. What is the electric potential at point A? a) V A < 0 b) V A = 0 c) V A > 0 A Question 2

+5 μC -3 μC Two charges q 1 = + 5 μC, q 2 = -3μC are placed at equal distances from the point A. What is the electric potential at point A? a) V A < 0 b) V A = 0 c) V A > 0 A Question 2 A positive test charge would move to the right

Question 3 Which of the following charge distributions produces V ( x ) = 0 for all points on the x -axis? (we are defining V ( x )  0 at x =  ) (a) x +2  C -2  C +1  C -1  C (b) x +2  C -1  C +1  C -2  C (c) x +2  C -1  C -2  C +1  C

Question 3 Which of the following charge distributions produces V ( x ) = 0 for all points on the x -axis? (we are defining V ( x )  0 at x =  ) (a) x +2  C -2  C +1  C -1  C (b) x +2  C -1  C +1  C -2  C (c) x +2  C -1  C -2  C +1  C The total potential at a point is the ALGEBRAIC sum of the contributions. Therefore, to make V ( x )=0 for all x, the + Q and - Q contributions must cancel. Thus any point on the x -axis must be equidistant from +2  C and -2  C and also from +1  C and -1  C. This condition is met only in case (a)!

Calculating Electric Potentials a b c uncharged conductor solid sphere with total charge Q I II III IV Calculate the potential V ( r ) at the point shown ( r < a ) r

Calculating Electric Potentials Calculate the potential V ( r ) at the point shown ( r < a ) Where do we know the potential, and where do we need to know it? Determine E ( r ) for all regions in between these two points Determine  V for each region by integration V =0 at r = ... we need r < a... a b c uncharged conductor solid sphere with total charge Q r I II III IV

Calculating Electric Potentials Calculate the potential V ( r ) at the point shown ( r < a )... and so on... a b c uncharged conductor solid sphere with total charge Q r I II III IV Check the sign of each potential difference  V (from the point of view of a positive charge)  V > 0 means we went “uphill”  V < 0 means we went “downhill”

Calculating Electric Potentials a b c r I II III IV Look at first term: Line integral from infinity to c has to be positive, pushing against a force: What’s left?

Calculating Electric Potentials a b c r I II III IV Look at third term : Line integral from b to a, again has to be positive, pushing against a force: What’s left? Previous slide we have calculated this already

Calculating Electric Potentials a b c r I II III IV Look at last term: Line integral from a to r, again has to be positive, pushing against a force. But this time the force doesn’t vary the same way, since “ r ’ ” determines the amount of source charge What’s left to do? ADD THEM ALL UP! Sum the potentials This is the charge that is inside “ r ” and sources field

Calculating Electric Potentials Calculate the potential V(r) at the point shown ( r < a ) a b c r I II III IV Add up the terms from I, III and IV: I III IV The potential difference from infinity to a if the conducting shell weren’t there An adjustment to account for the fact that the conductor is an equipotential,  V = 0 from c → b Potential increase from moving into the sphere

Calculating Electric Potentials Summary a b c r I II III IV The potential as a function of r for all 4 regions is: I r > c : II b < r < c: III a < r < b: IV r < a:

Let’s try some numbers a b c r I II III IV Q = 6  C a = 5cm b = 8cm c = 10cm I r > c : V ( r = 12cm) = kV II b < r < c: V ( r = 9cm) = kV III a < r < b: V ( r = 7cm) = kV IV r < a: V ( r = 3cm) = kV

Sparks High electric fields can ionize nonconducting materials (“dielectrics”) Breakdown can occur when the field is greater than the “dielectric strength” of the material. –E.g., in air, Insulator Conductor Dielectric Breakdown Ex. Arc discharge equalizes the potential What is ΔV? Note: High humidity can also bleed the charge off  reduce ΔV.

Question 3 Two charged balls are each at the same potential V. Ball 2 is twice as large as ball 1. As V is increased, which ball will induce breakdown first? (a) Ball 1 (b) Ball 2 (c) Same Time r1r1 Ball 1 r2r2 Ball 2

Question 3 Two charged balls are each at the same potential V. Ball 2 is twice as large as ball 1. As V is increased, which ball will induce breakdown first? (a) Ball 1 (b) Ball 2 (c) Same Time r1r1 Ball 1 r2r2 Ball 2  Smaller r  higher E  closer to breakdown Ex. High Voltage Terminals must be big!

Lightning! Factoids: _ + _ _ + + Collisions produce charged particles. The heavier particles (-) sit near the bottom of the cloud; the lighter particles (+) near the top. Stepped Leader Negatively charged electrons begin zigzagging downward. Attraction As the stepped leader nears the ground, it draws a streamer of positive charge upward. Flowing Charge As the leader and the streamer come together, powerful electric current begins flowing Contact! Intense wave of positive charge, a “return stroke,” travels upward at 10 8 m/s

Summary If we know the electric field E, allows us to calculate the potential function V everywhere (define V A = 0 above) Potential due to n charges: Equipotential surfaces are surfaces where the potential is constant.