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

you cant move and you feel like someones pushing you down onto your bed it can feel like you want to get up but you just can't it can be extremely scary

answer(20) psi;