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Voltage and Capacitance. Chapter 29. Electric Potential Energy. Potential Energy of a charge Wants to move when it has high PE Point b U = max K = min Point a U = min K = max. Electric Work. Charge moving between plates. Work = F D r cos0 o = qEx f - qEx i D U = qEx f - qEx i.
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Voltage and Capacitance Chapter 29
Electric Potential Energy Potential Energy of a charge • Wants to move when it has high PE • Point b • U = max • K = min • Point a • U = min • K = max
Electric Work Charge moving between plates. Work = FDr cos0o = qExf - qExi DU = qExf - qExi
A 2.0 cm X 2.0 cm capacitor has a 2.0 mm spacing and a charge of +1.0 nC. • Calculate the electric field of the capacitor. (2.82 X 105 N/C) • A proton is released from rest at the positive plate. Calculate the change in potential energy. (9.02 X 10-17 J) • Calculate the final speed of the proton. (3.29 X 105 m/s) • An electron is released from the halfway point between the plates. Calculate the change in PE and the final speed of the electron. (9.95 X 106 m/s)
A 2.0 cm diameter disk capacitor has a 2.5 mm spacing and a charge of +0.75 nC. • Calculate the electric field of the capacitor. • An electron is released from rest at the negative plate. Calculate the change in potential energy. (9.02 X 10-17 J) • Calculate the final speed of the electron. (3.29 X 105m/s)
Potential Energy of point charges Uelec = 1 q1q2 4peo r k = 9.0 X 109 Nm2/C2
A proton is fired from “far away” at a 1.0 cm diameter glass sphere of charge +100 nC. • Calculate the initial potential energy of the system (just as it touches the sphere). • Since PE = KE, calculate the needed initial speed of the proton.
An electron and a positron are created in the CERN collider. • Calculate the potential energy they have when they are 1.0 X 10-10 m apart. • Calculate the velocity they need to escape from one another. Remember that PE = KE, but you will need to consider the KE of both particles added together.
Electric Potential: Voltage • Voltage • 1 Volt = 1 Joule/Coulomb V = DU q Vab = Va – Vb = -Wba q Work done by the electric field to accelerate the charge
The higher the rock, the greater the PE • The greater the Voltage or charge, the greater the PE (DU = qV)
An electron is accelerated in a TV tube through a potential difference of 5000 V. • Calculate the change in PE of the electron (-8.0 X 10-16 J) • Calculate the final speed of the electron (m = 9.1 X 10-31 kg) (4.2 X 107 m/s (1/7th speed of light) Calculate the final speed of a proton (mass = 1.67 X 10-27 kg) (9.8 X 105 m/s (0.3% speed of light)
Equipotential Lines • Equipotential lines are perpendicular to electric field lines • Voltage is the same along equipotential lines • Like contour (elevation) lines on a map
Electric Field and Voltage U = qEs V = DU/q V = Es Greater the distance between plates, the greater the voltage The greater the E field, the greater the voltage
What is the electric field between two plates separated by 5.0 cm with a voltage of 50V. (1000 V/m)
A capacitor is constructed of 2.0 cm diameter disks separated by a 2.0 mm gap, and charged to 500 V. • Calculate the electric field strength (V = Ed) • Calculate the charge on each plate (E = Q/e0A) • A proton is shot through a hole in the negative plate towards the positive plate. It has an initial speed of 2.0 X 105 m/s. Does is have enough energy to reach the other side? (V = DPE/q)
Electron Volt • Energy an electron gains moving through a potential difference of 1 V 1 eV = 1.6 X 10-19 J • Ex: An electron moving through 1000 V would gain 1000 eV of energy
Voltage due to a Point Charge • Voltage is not directional (scalar) • Charged particles (i.e.: electrons, protons) have a voltage V = kQ r
Example 1 Consider a +1.0 nC charge. • Calculate the electric potential (voltage) at a point 1.0 cm from the charge • Calculate the electric potential at a point 3.0 cm from the charge.
Point Charges: Example 2 Calculate voltage (electric potential) at point A as shown below: A 30 cm 52 cm Q1 = +50 mC Q2 = -50 mC
Use Pythagoream theorem to calculate the distance from A to Q2: A 30 cm 52 cm Q1 = +50 mC Q2 = -50 mC
VA = V1 + V2 V1 = kQ= (9.00X 109 Nm2/C2)(5.00X10-5C) r (0.30 m) V1 = 1.50 X 106 V V1 = kQ= (9.00X 109 Nm2/C2)(-5.00X10-5C) r (0.60 m) V2 = -7.5 X 105 V VA = 1.50 X 106 V -7.5 X 105 V VA = 7.5 X 105 V
Point Charges: Example 3 Calculate voltage (electric potential) at point B as shown below: B 30 cm 26 cm 26 cm Q1 = +50 mC Q2 = -50 mC
Use Pythagoream theorem to calculate the distance from B to Q1and to Q2: B 30 cm 26 cm 26 cm Q1 = +50 mC Q2 = -50 mC
VA = V1 + V2 V1 = kQ= (9.00X 109 Nm2/C2)(5.00X10-5C) r (0.40 m) V1 = 1.125 X 106 V V1 = kQ= (9.00X 109 Nm2/C2)(-5.00X10-5C) r (0.40 m) V2 = -1.125 X 105 V VA = 1.125 X 106 V –1.125 X 105 V VA = 0 V
Point Charges: Example 4 How much work is required to bring a charge of q = 3.00 mC to a point 0.500 m from a charge Q = 20.0 mC? VQ = kQ= (9.00X 109 Nm2/C2)(2.00X10-5C) r (0.500 m) VQ = 3.6 X 105 V (This is the voltage caused by the stationary charge)
W = DU (like the work to lift a book to a shelf) V = DU q V = W q W = Vq W = (3.6 X 105 V )(3 X 10-6 C) W = 1.08 J
Point Charges: Example 5 Which of three sets of charges has the most: • positive potential energy? • The most negative potential energy? • Would require the most work to separate?
(i) (ii) (iii) - + - + + +
Charged Spheres Act as point charges At surface of sphere Substitute:
A proton is released from rest at the surface of a 1.00 cm-diameter sphere charged to +1000 V. • Calculate the charge on the sphere (0.56 nC) • Calculate the speed of the proton when it is 1.00 cm from the sphere (note that it has potential energy in both cases, U = qV) (3.6 X105 m/s)
A thin ring has a radius R and charge Q. Find the potential at a distance of z from the axis of the ring.
A thin disk has a radius R and charge Q. Find the potential at a distance of z from the axis of the ring.
A 17.5 mm diameter dime is charged to +5.00 nC. • Calculate the potential of the dime (10,300 V) • Calculate the potential 1.00 cm above the dime (3870 V)