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GPhys 122 electricity Electric Field MARLON FLORES SACEDON.

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Presentation on theme: "GPhys 122 electricity Electric Field MARLON FLORES SACEDON."β€” Presentation transcript:

1 gPhys 122 electricity Electric Field MARLON FLORES SACEDON

2 Electric Fields + What is electric field?
An electric field is a vector field that associates to each point in space the Coulomb force experienced by a unit electric charge. Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force it would exert on a positive test charge. The electric field is radially outward from a positive charge and radially in toward a negative point charge. +π‘ž +π‘ž +π‘ž +π‘ž +

3 Electric Fields What is electric field?
An electric field is a vector field that associates to each point in space the Coulomb force experienced by a unit electric charge. Electric fields converge and diverge at electric charges and they can be induced by time-varying magnetic fields. The electric field combines with the magnetic field to form the electromagnetic field.

4 Electric Fields Electric field lines
Outward direction of electric field Inward direction of electric field

5 Electric Fields

6 Electric Fields Electric field lines

7 Electric Fields - + 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2 𝐸= 𝐹 π‘ž 𝐹=π‘žπΈ 𝐸= π‘˜ π‘„π‘ž π‘Ÿ 2 π‘ž
+π‘ž 𝐹=π‘žπΈ π‘Ÿ 𝐸 𝐹 𝐸 𝐹 Definition of Electric field +π‘ž π‘Ÿ 𝐸= π‘˜ π‘„π‘ž π‘Ÿ 2 π‘ž - 𝑄 + 𝑄 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2 𝐸=π‘˜ 𝑄 π‘Ÿ 2 or Where: 𝐸 = electric field 𝑁 𝐢 , vector quantity π‘˜ = 9π‘₯ 𝑁. π‘š 2 𝐢 2 𝑄 = source charge 𝐢 +π‘ž = unit charge 𝐢

8 Electric Fields 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2
Problem: A point charge π‘ž=βˆ’8.0 𝑛𝐢 is located at the origin. Find the electric field vector at the field π‘₯=1.2π‘π‘š, 𝑦=βˆ’1.6 π‘π‘š. 𝐸=180,000 𝑁 𝐢 Magnitude of electric field 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2 +π‘₯ βˆ’π‘¦ _ π‘ž=βˆ’8.0 𝑛𝐢 πœƒ 1.2π‘π‘š 1.6 π‘π‘š 𝐸 π‘₯ =πΈπ‘π‘œπ‘ πœƒ π‘Ÿ 𝐸=π‘˜ 𝑄 π‘Ÿ 2 𝐸 π‘₯ =180,000 𝑁 𝐢 π‘π‘œπ‘  53.1 π‘œ 𝐸 𝐸 𝑦 π‘Ÿ= =2π‘π‘š =108, 𝑁 𝐢 πœƒ +π‘ž πœƒ= π‘‡π‘Žπ‘› βˆ’ = 53.1 π‘œ 𝐸 𝑦 =πΈπ‘ π‘–π‘›πœƒ 𝐸 π‘₯ 𝐸 𝑦 =180,000 𝑁 𝐢 𝑠𝑖𝑛 53.1 π‘œ 𝐸=9π‘₯ 𝑁. π‘š 2 𝐢 𝑛𝐢 2π‘π‘š 2 𝐸 = 𝐸 π‘₯ + 𝐸 𝑦 =143, 𝑁 𝐢 𝐸=9π‘₯ 𝑁. π‘š 2 𝐢 π‘₯ 10 βˆ’9 𝐢 2π‘₯ 10 βˆ’2 π‘š 2 𝐸 = 𝐸 π‘₯ βˆ’ 𝑖 + 𝐸 𝑦 + 𝑗 𝐸 =108, 𝑁 𝐢 βˆ’ 𝑖 +143, 𝑁 𝐢 + 𝑗 𝐸 =βˆ’ 108, 𝑁 𝐢 𝑖 βˆ’143, 𝑁 𝐢 𝑗

9 Electric Fields - - + 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2
Superposition of electric field - 𝑄 3 π‘Ÿ 3 𝐸= 1 4πœ‹ πœ– π‘œ βˆ™ 𝑄 π‘Ÿ 2 - 𝑄 2 𝐸=π‘˜ 𝑄 π‘Ÿ 2 or π‘Ÿ 2 𝐸 3 Where: 𝐸 = electric field 𝑁 𝐢 , vector quantity π‘˜ = 9π‘₯ 𝑁. π‘š 2 𝐢 2 𝑄 = source charge 𝐢 +π‘ž = unit charge 𝐢 𝐸 2 𝐸 1 + 𝑄 1 +π‘ž π‘Ÿ 1 𝐸 = 𝐸 𝐸 𝐸 3 +… Vector sum or Geometric sum

10 Electric Fields ASSIGNMENT 1: Find the electric field vector (Figure below) at the origin due to the surrounding charges 𝑄 1 , 𝑄 2 , & 𝑄 3 . Calculate also the magnitude and direction of electric field. 𝑦 π‘₯ + 𝑄 3 =4.0 πœ‡πΆ 5cm 3cm + 𝑄 2 =6.0 πœ‡πΆ 50 π‘œ - 𝑄 1 =βˆ’2.0 πœ‡πΆ 6cm

11 Electric Fields ASSIGNMENT 2:

12 Electric Fields ASSIGNMENT
3. At the three consecutive corners of a square 10 cm on the side are point charges of 50π‘₯ 10 βˆ’9 𝐢; 100π‘₯ 10 βˆ’9 𝐢; 100π‘₯ 10 βˆ’9 𝐢 respectively. Find the electrostatic field at the fourth corner of the square.

13 Electric Fields ASSIGNMENT 4. See the solutions below

14 Electric Fields ASSIGNMENT 5. See the solutions below

15 Electric Fields ASSIGNMENT 6. See the solutions below

16 Solutions to the problems

17 Solutions to the problems

18 Solutions to the problems

19 eNd


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