Lesson 17 Electric Fields and Potential. Eleanor Roosevelt High School Chin-Sung Lin. Electric Fields. Gravitational & Electric Forces. What are the formulas for the following physics laws?. Law of Universal Gravitation. Coulomb’s Law. Gravitational & Electric Forces. m 1 m 2. q 1 q 2.
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Gravitational Field • Gravitational Field: Force per unit mass Fg • Fg: gravitational force (N) • m: mass (kg) • g: gravitational field strength (N/kg, or m/s2) g = m
Electric Field • Electric Field: Force per unit charge Fe • Fe: electric force (N) • q: charge (C) • E: electric field strength (N/C) E = q
Electric Field Source Charge Test Charge +q +Q r Fe Fe E = q
Electric Field Source Charge Test Charge +q • –Q Fe r Fe E = q
Electric Field • Electric field is a vector • A vector includes ___________ and ____________ +q • –Q Fe Source Charge Test Charge r Fe E = +q +Q q r Fe
Electric Field • Electric field is a vector • A vector includes direction and magnitude +q • –Q Fe Source Charge Test Charge r Fe E = +q +Q q r Fe
Electric Field • Can you apply Coulomb’s law to this formula and then simplify it? Fe E = = ??? q Source Charge Test Charge +q +Q r Fe
Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) Fe q Q Q E = = k = k q r2q r2
Electric Field Example • What is the magnitude of the electric field strength when an electron experiences a 5.0N force?
Electric Field Example • What is the magnitude of the electric field strength when an electron experiences a 5.0N force? E = Fe / q E = 5 N / (1.6 x 10-19 C) = 3.13 x 1019 N/C
Electric Field Example • What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C?
Electric Field Example • What are the magnitude and direction of the electric field 1.5 m away from a positive charge of 2.1*10-9 C? E = kQ / r2 E = (8.99 x 109 N m2/C2) (2.1 x10-9 C) / (1.5 m)2 = 8.4 N/C Direction: away from the positive charge
Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle? (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle? [3 minutes] e – – 0.32 C 2.0 m
Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (a) What are the magnitude and direction of the electric field 2.0 m away from the particle? E = kQ / r2 E = (8.99 x 109 N m2/C2) (0.32 C) / (2.0 m)2 = 7.2 x 108 N/C Direction: toward the negative charge
Electric Field Exercise • There is a negative charged particle of 0.32 C in the free space. (b) What are the magnitude and direction of the electric force when an electron is placed 2.0 m away from this particle? E = Fe / q Fe = q E Fe = (1.6 x 10-19 C) (7.2 x 108 N/C) = 1.15 x 10-10 N
Aim: ElectricField • DoNow: (4 minutes) • Write down the definition of Electric Fieldin words • Write down theformulasof Electric Field in two different forms • Define every symbol in the formula and identify their units • Identify the relationships between Electric Field and other variables
Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) Fe Q E = = k q r2
Electric Field • Electric Field: Force per unit charge • Fe: electric force (N) • q: test charge (C) • Q: source charge (C) • E: electric field strength (N/C) • r: distance between charges (m) • k: electrostatic constant (N m2/C2) 1 1 E~ Q E~ Fe E ~ E ~ q r2
Electric Field • If you shift the test charge around, where can you find the electric field with the same magnitude? Source Charge Test Charge +q +Q Fe Fe E = q
Electric Field Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe
Electric Field • What will happen if you move the test charges away from the source charge? Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe
Electric Field Source Charge Test Charge +q +Q E Fe E Fe E Fe E Fe
Electric Field Source Charge Test Charge +q +Q E E Fe Fe E Fe E E Fe Fe E E Fe Fe E Fe
Electric Field Test Charge Source Charge Test Charge +q +Q +q E E Fe Fe E Fe E E Fe Fe E E Fe Fe E Fe
Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? (b) By how many electrons is the particle deficient?
Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (a) What is the charge on the particle? E = Fe / q Fe = E qFg = mg Fe = Fg E q = mg (1.2 x 107 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2) q = 4.80 x 10-16 C
Electric Field Example • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 1.2 x 107 N/C and are 2.00 mm apart. (b) By how many electrons is the particle deficient? e- = 1.6 x 10-19 C number of e- = 4.80 x 10-16 C / 1.6 x 10-19 C = 3000 e –
Electric Field Exercise • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle?
Electric Field Exercise • A charged droplet of mass 5.87 x 10-10 kg is hovering motionless between two parallel plates. The parallel plates have a electric field of 9.6 x 106 N/C and are 2.00 mm apart. What is the charge on the particle? E = Fe / q Fe = E qFg = mg Fe = Fg E q = mg (9.6 x 106 N/C) q = (5.87 x 10-10 kg) (9.81 m/s2) q = 6.0 x 10-16 C
Electric Field Exercise • A positively charged ball with mass 20 g is hanging between two charged parallel plates from the ceiling through an insulating wire with length 0.1 m. The electric field strength of the charged parallel plates is 4.2 x 109 N/C. When the ball is in balance, the wire and the vertical line form an angle of 60o. What is the charge of the ball?
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