Aparallel-plate capacitor is formed from two 7.2 cm-diameter electrodes spaced 1.8 mm apart. the electric field strength inside the capacitor is 3.0×106n/c. what is the magnitude of the charge (in nc) on each electrode?

432 nC

Explanation: In order to calculate the charge,  we know that the electric field corresponding for two parallel-plate is given by:

E=Q/(area*e0)

In our case E=3.0x10^6 N/C

Area = Pix (0,072 m)^2

e0 vacuum permittivity 8,85x 10^-12

i need with 1.3.2 investigate cycling o2 and co2 apex

we need to see it

96.77 amp

explanation:

rl = 250/100 = 2.5 ohms.

egl = 250 + (100 × 0.1) + 2 = 262 v.

at 700 r.p.m., eg2 = 262 × 700/1000 = 183.4 v.

if ia is the new current, then eg2 - 2 - (ia × 0.1) = 2.5 ia

which means this will give ia = 96.77 amp.