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.
F = qE & U = qV So, the relationship is: For the Electric Force: The General Relationship between a Conservative Force & its Potential Energy has the form: For the Electric Force: F = qE & U = qV So, the relationship is: Figure 23-5. To find Vba in a nonuniform electric field E, we integrate E·dl from point a to point b.
The simplest case is a Uniform 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 the electric field 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 the electric field 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.
Charged Conducting Sphere (a) r > r0 , (b) r = r0 , (c) r < r0 Example Charged Conducting Sphere 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, and compared with the electric field E. Use the relation: 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).
Example Charged Conducting Sphere Calculate the potential at a distance r from the sphere’s center for (a) r > r0 , (b) r = r0 , (c) r < r0 Use the relation: Vba = - [Q/(4πε0)] ∫[dr/(r2)] Limits ra = to rb 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).
Setting the potential V to zero at r = ∞ gives the General Form of the Potential Due to a Point Charge: 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 = qV = 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 = qV = 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)