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Electric Field (PHY N1203 - L01)

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• 1. LO1:Electric Force & Electric Field Lecture Slides Prepared by Physics Team
• 2. Learning Outcome (01) Define and solve problems related to electricity and the electric field.[ Chapter 23 & Chapter 24 ]  Sub-outcome 01: Define the electron, charge,     and Coulomb as a unit of electric charge. Sub-outcome 02: State Coulomb’s law and apply it to the solution of electric force problems. Sub-outcome 03: Define the electric field and explain what determines its magnitude and direction. Sub-outcome 04: Define the electric field intensity expression and compute its magnitude at a given distance from a known charge. Sub-outcome 05 : Explain Gauss’s Law, the concept of the permittivity of a medium, and the
• 3. The Electric Field 1. Now, consider point P a distance r from +Q. +q + P . 2. An electric field E exists at P if a test charge +q has a force F at that point. E r + ++ + Q ++ ++ 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: F Electric Field E F N ; Units q C
• 4. 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.
• 5. 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.
• 6. 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 E F ; Units q N C The direction of E at a point is the same as the direction that a positive charge would move IF placed at that point.
• 7. Example 1. A +2 nC charge is placed at a distance r from a –8 C charge. If the +2 nC charge experiences a force of 4000 N, . +q + P what is the electric field intensity E at 4000 N point P? E E r First, we note that the direction of E is toward –Q (down). - -Q- - –8 C -- Electric Field E F q 4000 N -9 2 x 10 C E = 2 x 1012 N/C Downward Note: The field E would be the same for any charge placed at point P. It is a property of that space.
• 8. Example 2. 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 . The E-field is downward, and the force on e- is up. E F ; F q F qE + + + + + + + + + e- - qE Fe. E e- - - - - - - - - - -19 (1.6 x 10 C)(4 x 10 F = 6.40 x 10-15 N, Upward 4 N C )
• 9. 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. E F . +q +. P P r The outward force on +q is: ++ ++ E +Q + ++ + kQq 2 r F The electric field E is therefore: E F q kQq r q 2 E kQ 2 r kQ 2 r
• 10. Example 3. What is the electric field intensity E at point P, a distance of 3 m from a negative charge of –8 nC? First, find the magnitude: E=? . 9 Nm2 (9 x 10 C2 )(8 x 10-9C) kQ P r E 2 2 r (3 m) 3m -Q -8 nC 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
• 11. 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: E kQ r2 q1 ER E2 A E1 q3 - E3 + q2 Directions are based on positive test charge.
• 12. Example 4. Find the resultant field at point A due to the –3 nC charge and the +6 nC charge arranged as shown. E for each q is shown with direction given. -3 nC q1 - 3 cm E 5 cm 1 E2 A 4 cm E1 9 Nm2 C2 (9 x 10 +6 nC + q2 -9 )(3 x 10 C) (3 m) E1 2 E2 kq1 ; E2 2 r1 9 Nm2 C2 (9 x 10 kq2 r22 )(6 x 10-9C) (4 m)2 Signs of the charges are used only to find direction of E
• 13. Example 4. (Cont.) Find the resultant field at point A. The magnitudes are: -3 nC q1 - 3 cm E 5 cm 1 E1 +6 nC E2 A 4 cm E1 = 3.00 N, West + q2 E2 9 Nm2 C2 (9 x 10 (3 m)2 9 Nm2 C2 (9 x 10 2 E2 E2 = 3.38 N, North R12 ; tan )(6 x 10-9C) (4 m)2 Next, we find vector resultant ER ER )(3 x 10-9C) E1 E2 ER E1 E2
• 14. Example 4. (Cont.) Find the resultant field at point A using vector mathematics. ER E1 = 3.00 N, West E1 E2 E E2 = 3.38 N, North Find vector resultant ER (3.00 N) 2 (3.38 N) 2 4.52 N; = 48.40 N of W; or tan 3.38 N 3.00 N = 131.60 Resultant Field: ER = 4.52 N; 131.60
• 15. Electric Field Lines Electric Field Lines are imaginary lines drawn in such a way that their direction at any point is the same as the direction of the field at that point. + ++ + Q ++ ++ - -- -Q -- Field lines go away from positive charges and toward negative charges.
• 16. Examples of E-Field Lines Two equal but opposite charges. Two identical charges (both +). Notice that lines leave + charges and enter - charges. Also, E is strongest where field lines are most dense.
• 17. The Density of Field Lines Gauss’s Law: The field E at any point in space is proportional to the line density at that point. Gaussian Surface N Line density r A Radius r N A
• 18. Line Density and Spacing Constant Consider the field near a positive point charge q: Then, imagine a surface (radius r) surrounding q. Radius r r E is proportional to N/ A and is equal to kq/r2 at any point. N A Gaussian Surface N A Define 0 E Where E; kq 2 r E as spacing constant. Then: 0 is: 1 0 4 k
• 19. Permittivity of Free Space The proportionality constant for line density is known as the permittivity and it is defined by: 1 0 4 k 8.85 x 10-12 C2 2 N m Recalling the relationship with line density, we have: N A 0 E or N Summing over entire area A gives the total lines as: 0 E A N= oEA
• 20. Example 5. Write an equation for finding the total number of lines N leaving a single positive charge q. Draw spherical Gaussian surface: Radius r N r 0 E A and N 0 EA Substitute for E and A from: E Gaussian Surface N 0 EA 0 kq r2 q ; A = 4 r2 4 r2 q (4 r 2 ) 4 r2 N= oqA =q Total number of lines is equal to the enclosed charge q.
• 21. Gauss’s Law Gauss’s Law: The net number of electric field lines crossing any closed surface in an outward direction is numerically equal to the net total charge within that surface. N 0 EA If we represent q as net enclosed positive charge, we can write rewrite Gauss’s law as: q EA q 0
• 22. Example 6. How many electric field lines pass through the Gaussian surface drawn below. First we find the NET charge q enclosed by the surface: q = (+8 –4 – 1) = +3 C N 0 EA q N = +3 C = +3 x 10-6 lines Gaussian surface -4 C q1 -1 C q3 - +8 C q2 + q4 + +5 C
• 23. Example 7. A solid sphere (R = 6 cm) having net charge +8 C is inside a hollow shell (R = 8 cm) having a net charge of –6 C. What is the electric field at a distance of 12 cm from the center of the solid sphere? Draw Gaussian sphere at radius of 12 cm to find E. N 0 EA q q = (+8 – 6) = +2 C q qnet ; E 0 AE 0A E q (4 r 2 ) 0 Gaussian surface -6 C 8cm - - 12 cm - +8 C - 6 cm - - 2 x 10-6C -12 Nm2 2 (8.85 x 10 2 )(4 )(0.12 m) C
• 24. Example 7 (Cont.) What is the electric field at a distance of 12 cm from the center of the solid sphere? Draw Gaussian sphere at radius of 12 cm to find E. N 0 EA q Gaussian surface q = (+8 – 6) = +2 C q qnet ; E 0 AE 0A E - 8cm 2 C 1.25 x 106 N C 2 0 (4 r ) 12 cm +8 C -6 C - 6 cm - - E = 1.25 MN/C
• 25. Summary of Formulas The Electric Field Intensity E: The Electric Field Near several charges: Gauss’s Law for Charge distributions. E F q 0 N Units are C kQ r2 E kQ r2 Vector Sum EA q; q A