Learning Goal Coulomb S Law Electric Fields And The Connection Between The Electric Field And The Electric Force Cou 1 (128.33 KiB) Viewed 21 times
Learning Goal: Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force Facting between two charges. The magnitude F of the force between two charges 6.00 μC and 3.00 μC depends on the product of the charges and the square of the distance r between the charges: F = k 91 92 p2 where k = 1/(ATED) = 8.99 × 10⁰ N·m²/C². The direction of the Figure < 1 of 2 Electric Field Vector due to (created by) a single point charge Q Magnitude: E = -=k Q=4.00μC r=0.7m E-k-(8.99x10' N-m²/C¹) (4.0x10-C) -7.34×10* N/C (0.70) E direction E You will use a new approach to solve the previous problem in which three point electric charges are on a straight line. (Figure 2)on the left panel shows ithe configuration for this problem.. Before putting anyt charge at location A: q2 = 3.00 μC , q3 = -4.00 μC 1 μC = 106 c d12 = 0.20 m, d23 = 0.50 m use k = 9.00×109 N/(m² C²) for the Coulomb Constant. ▶ Part A - Before putting anyt charge at location A: Calculate the magnitude of the electric FIELD at location created by charge q2. Part B - What is the direction of the electric FIELD at location A created by charge q2? ▶ Part C - Before putting anyt charge at location A: Calculate the magnitude of the electric FIELD at location A created by charge q3. Part D - What is the direction of the electric FIELD at location A created by charge q3? ▶ Part E - Before putting any charge at location A, what is the NET electric FIELD (including both magnitude and direction) at location A (by charge q2 and q3? ▶ Part F - Case 1: Put a POSITIVE charge q₁ = 6.00 μC at location A. What is the magnittude of the NET electric FORCE ON charge q1 BY charge q2 and q3?
Learning Goal: Coulomb's law, electric fields, and the connection between the electric field and the electric force. Coulomb's law gives the electrostatic force Facting between two charges. The magnitude F of the force between two charges 6.00 μC and 3.00 μC depends on the product of the charges and the square of the distance r between the charges: F = k 91 92 p2 where k = 1/(ATTER) = 8 99 × 10⁰ N·m²/C². The direction of the Figure < 1 of 2 Electric Field Vector due to (created by) a single point charge Q Magnitude: E = -=k Q=4.00μC r=0.7m E-k-(8.99x10' N-m²/C¹) (4.0x10-C) -7.34×10* N/C (0.70) E direction E Part E - Before putting any charge at location A, what is the NET electric FIELD (including both magnitude and direction) at location A (by charge q2 and q3? Part F - Case 1: Put a POSITIVE charge q₁ = 6.00 μC at location A. What is the magnittude of the NET electric FORCE ON charge q1 BY charge q2 and q3? ▶ Part G - Case 1: Put a POSITIVE charge q1 at location A. What is the NET electric FORCE (including magntide and direction) ON charge q1 BY charge q2 and q3? q1 in case 1 is removed from location A. Case 2: q₁/= -6.00x10-6 C is put at location A. ▶ Part H - Case 2: What is the NET electric FORCE (including magnitude and direction) on charge q1' (by charge q2 and q3)? Compare this NEW approach (Two stages -- Net Electric Field, then Net Electric Force) to previous approach (One stage - Net Electric Force). What is the advantage of this NEW approcach (even though it has TWO stages)? INSIGHT of this new approach: When different q1's are placed at location A, you DON"T need to redo the calculation of the NET electric field at location A (created by q2 and 93). Provide Feedback Next >
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