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Net Electric Field due to Two Point Charges (13:56)

Lecture Notes
AP Physics 2
APC E&M
Bobby and Bo solve a two-point-charge problem step by step: with q₁ = +3.0 μC at the origin and q₂ = -6.0 μC at x = 0.40 m, the class determines the net electric field at point P, located 0.30 m in the positive y-direction. After converting both charges from microcoulombs to coulombs, Billy explains why electric field is a vector, and the group reasons out the direction of each individual field using the rule that a field points away from a positive charge and toward a negative one. Bobby calculates the electric field from charge 1 using E = kQ/r² (300,000 N/C), then Bo uses the Pythagorean theorem on the resulting 3-4-5 right triangle to find the distance from charge 2 to point P (0.5 m) before finding the magnitude of the electric field from charge 2 (216,000 N/C). Since that field isn't purely in the x- or y-direction, the class resolves it into components using sine and cosine, then sums the electric fields in each direction: 172,800 N/C in the x-direction and 170,400 N/C in the y-direction. Using the Pythagorean theorem and an inverse tangent one final time, they find the net electric field at point P: 240,000 N/C at 45° above the positive x-direction. Billy connects the whole process back to the Five Steps to Help Solve any Free Body Diagram, since summing electric field vectors follows the exact same pattern as summing force vectors.

Chapters:
0:00 The Problem and Converting Charges to Coulombs
2:09 Determining the Directions of the Electric Field Vectors
3:39 Magnitude of the Electric Field from Charge 1
4:49 Distance to Charge 2 and Magnitude of the Electric Field from Charge 2
6:03 Resolving the Electric Field from Charge 2 into Components
9:03 Redrawing the Diagram and Summing the Components
10:17 Solving for the Net Electric Field and the Free Body Diagram Connection
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