Question

Two 3.0 mm × 3.0 mm electrodes are held 0.10 mm apart and are attached to 9.0 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?
The constant K will be 3.1

Answer 1

a) 9v

b) 90kv/m

c) 7.17pC

d) 9v

e) 29kv/m

f) 22.3pC

Step-by-step explanation:

Capacitor's potential difference before the Mylar is inserted is

- The potential difference across the two plates is same as the voltage provided by the battery V = 9V which remains constant throughout.

The capacitor's electric field before the Mylar is inserted is, E = v/kd

E = 9 / 1*10^-4

E = 90000v/m

The capacitor's charge before the Mylar is inserted is, C = k*A*ε / d

C = 9*10^-6 * 8.85*10^-12 / 1*10^-4

C = 7.965*10^-13

C = 0.7965pF

Q = CV, 0.7965 * 9

Q = 7.17pC

Capacitor's potential difference after the Mylar is inserted, as stated earlier is constant at v = 9v

Capacitor's electric field after the Mylar is inserted, E = v/kd

E = 9/1*10^-4 * 3.1

E = 9/3.1*10^-4

E = 29032v/m

Capacitor's charge after the Mylar is inserted will be, C = k*A*ε / d

C = (3.1 * 9*10^-6 * 8.85*10^-12) / 1*10^-4

C = 2.47*10^-16 / 1*10^-4

C = 2.47*10-12

C = 2.47pF

Q = CV, = 9 * 2.47 = 22.3pC