Determine the separation between the plates

[joemama69]

Homework Statement

As shown in the circuit, a potential difference of 625 V is applied across an air-filled parallel plate capacitor having capacitance c1 = 2.36X10^-11 F and plate area A = 4X10^-3 m^2. the dielectric constanf of air is 1. consider the switch in the circuit to be opened as show for parts a & b

a) find teh charge of capacitor c1

b) determine the separation between the plates at c1

capacitor c2 is uncharged when the switch is closed. capacitor c2 is paralled plate capactiro haveing the same plate area and separation as c1. c2 is filled with polystyrene having a dielectric constant 2.56

c) how much charge passes throught the switch after it is closed

d) what is the equivalat capacitance of the two capacitors in the circuit

Homework Equations

The Attempt at a Solution

a) find Q at c1

Q = CV = 2.36X10^-11 * 625 = 1.475X10^-8 J


b) find distance of c1 plates

C = E_o*A/d = 8.85X10^-12 * 4X10-3/d = 2.36X10^-11 therefore d = 1.5X10^-3 m


c) c2 = Kc1 = 2.56 * 2.36X10^-11 = 6.04X10^-11

C_eq = c1 + c2 = 8.4X10^-11

Q = c_eq * V = 8.4X10^-11 * 625 = 5.25X10^-8 J


d) fine C_eq for the capacitors


C_eq was determined in part c which makes me questions part c.

 

[AvroArrow]

if C_eq is 8.4X10^-11 then the charge passing through the switch after it is closed should be 8.4X10^-11 * 625 = 5.25X10^-8 J