Question

A parallel-plate capacitor filled with air has plates of area 7.0×10⁻³ m2m2 and a separation of 0.32 mmmm .

A) Find the magnitude of the charge on each plate when the capacitor is connected to a 12-VV battery.

b) Will your answer to part A increase, decrease, or stay the same if the separation between the plates is increased?

c) Explain.

d) Calculate the magnitude of the charge on the plates if the separation is 0.90 mmmm .

Answer 1

We know that fora a parallal plate capacitors CA _value of capacitancesco (1) C = Permittivity of dielectric A = Area of plate overlapio square __meters. dl= Distance beboeen plates in meters Va Voltage Changes g=CV_-(1) poom- (1) CE CA. 148.85X10 ** 7.0X10 3- 0.032x10-2 here for aire value of E= 8.85x60"? F/m. Area, A=7.0810 m² Seperation, d = 32 mm=0.32X163 m. d AD_ from, cis C = 1935X20-35 pe 1935x10143 x 12 = 23230x10" _-_23.2 x 10°C LB &cIt the seperation between plaket will increase, then the value of capacitance will be decreased, as, Cafe as a result, the value of charge will be decreassed, as sac. So, the charge stored inside the capacitor will be decreased. a _ for, de 0.90 mm = 0.90x10*3m C= EA. 8.85x1072 x 7.0 x10-3 0.90 X10-3 -68083x10*12 = 688.3 x1013.00 CV = 688.3x10mx12 Farad = 8259.6x1099 = 8.259 XI0