Relation Between Electric Potential V & Electric Field E Figure 23-5. To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.

Work = Charge  Potential Difference: Also, Work = Force  Displacement: Equate the expressions for W & solve for E (assumed constant),

The simplest case is a Uniform (constant) E Field

E field obtained from voltage Example E field obtained from voltage Two parallel plates are charged to produce a potential difference of 50 V. The plate separation is d = 0.050 m. Calculate the magnitude of E in the space between the plates. Solution: E = V/d = 1000 V/m.

E field obtained from voltage Example E field obtained from voltage Two parallel plates are charged to produce a potential difference of 50 V. The plate separation is d = 0.050 m. Calculate the magnitude of E in the space between the plates. E = (Vba)/d = (50)/(0.05) E = 1,000 V/m Solution: E = V/d = 1000 V/m.

Electric Potential Due to Point Charges The electric potential due to a point charge can be derived from the field using calculus. It has the form:

These plots show the potential due to (a) positive and (b) negative charge. Using potentials instead of fields can make solving problems easier. The potential is a scalar quantity, but the field is a vector.

Electric Field Near a Metal Sphere A solid metal sphere carries an excess positive charge, q As previously discussed, from Gauss’s Law: The excess charge is on the surface. The field inside the metal is zero (Static conditions only!)

Electric Field Near a Metal Sphere The field outside any spherical ball of charge is: r = distance from the center of the ball This holds outside the sphere only, the field is still zero inside the sphere

Electric Potential Near a Metal Sphere Since the field outside the sphere is the same as that of a point charge, the potential is also the same So, the potential is constant inside the sphere

The potential at all locations in the metal sphere must equal the potential at the outer edge of the sphere. So, everywhere inside a metal sphere is:

Charged Conducting Sphere (a) r > r0 , (b) r = r0 , (c) r < R0 Example Charged Conducting Sphere (r0  rsphere) A conducting sphere is uniformly charged with total charge q. Calculate the potential at a distance r from the sphere’s center for (a) r > r0 , (b) r = r0 , (c) r < R0 V is plotted here, & compared with the electric field E. Use the relations: Solution: The electric field outside a conducting sphere is Q/(4πε0r2). Integrating to find the potential, and choosing V = 0 at r = ∞: a. V = Q/4πε0r. b. V = Q/4πε0r0. c. V = Q/4πε0r0 (the potential is constant, as there is no field inside the sphere).

General Form of the Potential Due to a Point Setting the potential V to zero at r = ∞ gives the General Form of the Potential Due to a Point Charge or a Conducting Sphere: Figure 23-10. Potential V as a function of distance r from a single point charge Q when the charge is positive. Figure 23-11. Potential V as a function of distance r from a single point charge Q when the charge is negative.

Example: Work required to bring two positive charges close together: Calculate the minimum work that must be done by an external force to bring a charge q = 3 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20 µC. Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

W = qV = q(Vb –Va) = qke[(Q/rb) – keQ(Q/ra)] 1.08 J Example: Work required to bring two positive charges close together: Calculate the minimum work that must be done by an external force to bring a charge q = 3 μC from a great distance away (take r = ∞) to a point 0.500 m from a charge Q = 20 µC. W = qV = q(Vb –Va) = qke[(Q/rb) – keQ(Q/ra)] 1.08 J Solution: The work is equal to the change in potential energy; W = 1.08 J. Note that the field, and therefore the force, is not constant.

Example: Potential above two charges (a) Calculate the electric potential at point A in the figure due to the two charges shown. (b) Repeat the calculation for point B. (Solution on white board). Solution: The total potential is the sum of the potential due to each charge; potential is a scalar, so there is no direction involved, but we do have to keep track of the signs. a. V = 7.5 x 105 V b. V = 0 (true at any point along the perpendicular bisector)

Cathode Ray Tube: TV & Computer Monitors, Oscilloscope Some Applications Cathode Ray Tube: TV & Computer Monitors, Oscilloscope A cathode ray tube contains a wire cathode that, when heated, emits electrons. A voltage source causes the electrons to travel to the anode. Figure 23-21. If the cathode inside the evacuated glass tube is heated to glowing, negatively charged “cathode rays” (electrons) are “boiled off” and flow across to the anode (+) to which they are attracted.

The electrons can be steered using electric or magnetic fields. Figure 23-22. A cathode ray tube. Magnetic deflection coils are often used in place of the electric deflection plates shown here. The relative positions of the elements have been exaggerated for clarity.

Old fashioned televisions and computer monitors (not LCD or plasma models) have a large cathode ray tube as their display. Variations in the field steer the electrons on their way to the screen. Figure 23-23. Electron beam sweeps across a television screen in a succession of horizontal lines. Each horizontal sweep is made by varying the voltage on the horizontal deflection plates. Then the electron beam is moved down a short distance by a change in voltage on the vertical deflection plates, and the process is repeated.

An oscilloscope displays an electrical signal on a screen, using it to deflect the beam vertically while it sweeps horizontally. Figure 23-24. An electrocardiogram (ECG) trace displayed on a CRT.