Question

# Two 10.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal...

Two 10.0-cm-diameter electrodes 0.59 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 Vbattery.

Part A

What is the charge on each electrode while the capacitor is attached to the battery?

q1, q2 =   C

Part B

What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

E =

Part C

What is the potential difference between the electrodes while the capacitor is attached to the battery?

V| =

Part D

What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

q1, q2 =   C

Part E

What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

E =

Part F

What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.1 cm apart? The electrodes remain connected to the battery during this process.

V| =

given

distance between the plates, d = 0.59 cm = 0.0059 m

area of the plates, A = pi*0.1^2/4 = 0.00785 m^2

V = 14 V

A) Capacitance of the capacitor, C = A*epsilon/d

= 0.00785*8.854*10^-12/(0.0059)

= 1.17*10^-11 F

so, charge on capacitor pates, q = C*V

= 1.17*10^-11*14

= 1.64*10^-10 C

q1, q2 = 1.64*10^-10 , -1.64*10^-10   C

B)

E = V/d

= 14/0.0059

= 2373 N/c

C) delta_V = 14 V

D) C = Capacitance of the capacitor, C = A*epsilon/d

= 0.00785*8.854*10^-12/(0.011)

= 6.32*10^-12 F

so, charge on capacitor pates, q = C*V

= 6.32*10^-12*14

= 8.85*10^-11 C

q1, q2 = 8.85*10^-11 , -8.85*10^-11 C

E) E = V/d

= 14/0.011

= 1273 N/C

F) delta_V = 14 V

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