Question

Two 9.00 µC charged particles are located on the x axis. One is at x = 1.00 m, and the other is at x = -1.00 m.
(a) Determine the electric field on the y axis at y = 0.900 m.

b) Calculate the electric force on a -3.00 µC charge placed on the y axis at y = 0.900 m.

Answer 1

field due to either charge

E = k q / r^2 = 9* 10^9* 9* 10^-6 / ( 1^2 + 0.9^2)

E = 4.26* 10^5 N/C

angle made by E with vertical

x = arctan ( 1/ 0.9) = 48.01

net field at given point

En^2 = E^2 + E^2 + 2* E^2 cos 48.01

En = 7.783* 10^5 N/C

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b)

F = qE = 2.335 N

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