Electric Field Electric field a region of space in which electric charges will feel a force The Concept of a Field A field is defined as a property of space in which a material object experiences a force. Above earth, we say there is a gravitational field at P. m . P F Because a mass m experiences a downward force at that point. No force, no field; No field, no force! The direction of the field is determined by the force. The Gravitational Field A F B F If g is known at every point above the earth then the force F on a given mass can be found. Consider A and above Note that points the force F isBreal, but the field surface of the earth—justway is just a convenient points in space. of describing space. The field at points A or B might be found from: F g m The magnitude and direction of the field g depends on the weight, which is the force F. Representing electric field • Electric field is represented by field lines (lines of force) Tells us two things: 1. direction of field 2. strength of field Representing electric field 1. Radial field spreads out in all directions a. Positive charge (outward) b. Negative charge (inward) Representing electric field 2. Electric field of two oppositely charged objects Representing electric field 3. Electric field of two like charged objects Representing electric field 4. Uniform electric field has the same strength at all points and produced between oppositely charged parallel plates Electric field lines always begin on a positive charge and end on a negative charge and do not stop in midspace. The Electric Field 1. Now, consider point P a distance r from +Q. 2. An electric field E exists at P if a test charge +q has a force F at that point. 3. The direction of the E is the same as the direction of a force on + (pos) charge. 4. The magnitude of E is given by the formula: +q P +. F E r + ++ + ++Q++ Electric Field F N E ; Units q0 C Field is Property of Space F . +q + E r ++ + + ++Q++ Electric Field Force on +q is with field direction. Force on -q is against field direction. -q -. F E r ++ + + ++Q++ Electric Field The field E at a point exists whether there is a charge at that point or not. The direction of the field is away from the +Q charge. Field Near a Negative Charge E +q +. F r - --- -Q -Electric Field Force on +q is with field direction. Force on -q is against field direction. F -q -. E r -- -Q --Electric Field Note that the field E in the vicinity of a negative charge –Q is toward the charge—the direction that a +q test charge would move. The Magnitude of E-Field The magnitude of the electric field intensity at a point in space is defined as the force per unit charge (N/C) that would be experienced by any test charge placed at that point. Electric Field Intensity E F N E ; Units q0 C q0 - test charge (has a small magnitude so it does not affect the other charges) Example 1. The positive test charge has a magnitude of 3.0x10-8C and experiences a force of 6.0x10-8N. (a)Find the force per coulomb that the test charge experiences. (b)Predict the force that a charge of +12x10-8C would experience if it replaced the test charge. (a) F 6.0 10 8 N 2.0 N C 8 qo 3.0 10 C (b) F 2.0 N C 12.0 108 C 24 108 N Example 2. The charges on the two metal spheres and the ebonite rod create an electric field at the spot indicated. The field has a magnitude of 2.0 N/C. Determine the force on the charges in (a) and (b) (a) F qo E 2.0 N C 18.0 108 C 36 108 N (b) F qo E 2.0 N C 24.0 108 C 48 108 N The E-Field at a distance r from a single charge Q Consider a test charge +q placed at P a distance r from Q. The outward force on +q is: kQq0 F 2 r FE +q +.. P P r kQ ++ E 2 + + + r +Q + ++ The electric field E is therefore: F kQq0 r 2 E q0 q0 kQ E 2 r Example 3. What is the electric field intensity E at point P, a distance of 3 m from a negative charge of –8 nC? . E=? r 3m -Q -8 nC First, find the magnitude: P 9 Nm 2 C2 -9 )(8 x 10 C) kQ (9 x 10 E 2 r (3 m) 2 E = 8.00 N/C The direction is the same as the force on a positive charge if it were placed at the point P: toward –Q. E = 8.00 N, toward -Q Example 4: The Electric field of a Point Charge The isolated point charge of q=+15μC is in a vacuum. The test charge is 0.20m to the right and has a charge qo=+0.80μC. Determine the electric field at point P. The Electric Field F k q qo r2 8.99 10 9 N m 2 C 2 0.80 10 6 C 15 10 6 C 0.20m 2 F 2.7 N 6 E 3 . 4 10 NC -6 qo 0.80 10 C 2.7 N Example 5: The Electric Fields from Separate Charges May Cancel Two positive point charges, q1=+16μC and q2=+4.0μC are separated in a vacuum by a distance of 3.0m. Find the spot on the line between the charges where the net electric field is zero. Ek q r2 Example 6. A constant E field of 40,000 N/C is maintained between the two parallel plates. What are the magnitude and direction of the force on an electron that passes horizontally between the plates.c + + + + + + + + + The E-field is downward, and the force on e- is up. e- - F E ; F qE q Fe -. E e- - - - - - - - - - - F qE (1.6 x 10 C)(4 x 10 -19 F = 6.40 x 10-15 N, Upward 4 N C ) The Resultant Electric Field The resultant field E in the vicinity of a number of point charges is equal to the vector sum of the fields due to each charge taken individually. Consider E for each charge. Vector Sum: E = E1 + E2 + E3 Magnitudes are from: kQ E 2 r q1 ER E2 A E1 q3 - E3 + q2 Directions are based on positive test charge. Example 7. Find the resultant field at point A due to the –3 nC charge and the +6 nC charge arranged as shown. -3 nC q1 3m E1 5m +6 nC E2 A 4 m E1 (9 x 10 E for each q is shown with direction given. 9 Nm 2 C2 + q2 -9 )(3 x 10 C) (3 m) kq1 kq2 E1 2 ; E2 2 r1 r2 2 E2 (9 x 10 9 Nm 2 C2 )(6 x 10-9C) (4 m) 2 Signs of the charges are used only to find direction of E Example 7. (Cont.)Find the resultant field at point A. The magnitudes are: -3 nC q1 3m E1 E1 5m +6 nC E2 A 4 m + q2 E2 (9 x 10 9 Nm 2 C2 (3 m) 2 (9 x 10 9 Nm 2 C2 E1 = 3.00 N, North E2 = 3.38 N, West Next, we find vector resultant ER E1 ER E R ; tan E2 2 2 2 1 )(3 x 10-9 C) )(6 x 10-9C) (4 m) 2 ER E2 E1 Example 7. (Cont.)Find the resultant field at point A using vector mathematics. ER E1 = 3.00 N, North E2 E1 E2 = 3.38 N, West Find vector resultant ER E (3.00 N) 2 (3.38 N) 2 4.52 N; 3.00 N tan ∅ = 3.38 N = 41.60 N of W; or q = 48.40 W of N Resultant Field: ER = 4.52 N at 138.41°