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mixer [17]
2 years ago
8

HELP PLEASE!!! In the diagram, q1 = -6.39*10^-9 C and q2 = +3.22*10^-9 C. What is the electric field at point P? Include a + or

- sign to indicate the direction. 91 92 P 0.150 m ** 0.250 m (Hint: The distance from q1 to P is NOT 0.150 m.) (Remember, E points away from + charges, and toward charges.) (Unit = N/C) HELP PLEASE!!​

Physics
1 answer:
horrorfan [7]2 years ago
8 0

Answer:

E = K Q / R^2

E1 = K Q1 / R1^2 = K * (-6.39 E-9) / .4^2 = -K 39.9 E-9

E2 = K Q2 / R2^2 = K * 3.22 E-9 / .25^2 = K 51.5 E-9

Field will point to right of P since E2 (positive) stronger

E = K (E2 - E1) = 9.0 E9 * (51.5 - 39.9) E-9 = 104 N/C

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