Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.10-mm-thick sheet of Mylar is inserted between the electrodes. A) What is the capacitor's potential difference before the Mylar is inserted?
B) What is the capacitor's electric field before the Mylar is inserted?
C) What is the capacitor's charge before the Mylar is inserted?
D) What is the capacitor's potential difference after the Mylar is inserted?
E) What is the capacitor's electric field after the Mylar is inserted?
F) What is the capacitor's charge after the Mylar is inserted?

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

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