Calculate the capacitance of a capacitor when each plate has an area of 1.2 cm2 and are 2.5 mm apart. 2. calculate the maximum charge that each plate above can hold if a voltage difference of 12.0 v is applied to the capacitor. 3. calculate the magnitude of the uniform electric field produced by this capacitor if the voltage supplied is 12.0 v

a) 0.43 pF (10~-12F);  5.16 pC (10^-12C); 4.85 *10^3 N/C

Explanation: In order to solve this problem we have to consider the expresion  for the capacitor of parallel plates, which is

ΔV=Q/C in this case C=A*εo/d where A is the area of the plates and d its separation,

The we have;

C=0,00012 m^2* 8,85*10~ -12/0.0025 m=  0.43 pF

Then we have that

Q= ΔV*C= 12V * 0.43 pF=5.16 pC

Finally, the electric field inside the plates is given by:

E=Q/(A*εo)= 5.16 pC/(0.00012 m^2*8.85 *^10-12 C^2/N*m^2)=4.85 *10^3 N/C

one person pulls 90n one way and the other has 100n pulling the other way.

so the net force = 90 - 100 = -10n,

or the ans can be 10n if the picture shows -90n and 100n instead.

its a circle not a square that's simple silly.

explanation: