I'LL GIVE ONE OF YOU BRAINLIEST ANSWER. PLEASE HELP ME
a) The electric field between two equally but oppositely charged parallel plates is 2.35 N/C [south]. If a charge of –2.00 C was placed at the midway point between the plates, what force (magnitude and direction) would act on the charge?
Give the direction as acting north or south. b) For another set of parallel plates, a charge of –4.00 C experiences a force of 6.00 x 10–3 N [east]. Determine the magnitude and direction of the electric field between these plates. Give the direction as east or west. (continued) 58 G r a d e 1 2 P h y s i c s Assignment 7.4: Electric Field between the Plates of a Parallel Plate Capacitor (continued) d) Label each plate in the diagram below with the appropriate polarity according to the information from part (c) above

a) The magnitude of the electric field = (electric force)/(charge of particle)

Electric force = (charge of particle) (electric field magnitude)
= (2 C)(2.35 N/C) = 4.7 N

(electric force and electric field are always in the same direction)
so, force = 4.7 N south

b) Electric field magnitude = (electric force)/(particle charge)
= (6*10^-3 N)/(-4 C) = -0.0015 N/C = -1.5*10^-3 N/C east

For part d, I think I'll need to see part c first.

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