Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. T

Question

Two 13.0-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14 V battery. Part A. What is the charge on each electrode while the capacitor is attached to the battery? Enter your answers numerically separated by a comma. Part B. What is the electric field strength inside the capacitor while the capacitor is attached to the battery? Express your answer with the appropriate units. Part C. What is the potential difference between the electrodes while the capacitor is attached to the battery? Express your answer with the appropriate units. 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 1.3 cm apart? The electrodes remain connected to the battery during this process. Enter your answers numerically separated by a comma. 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 1.3 cm apart? The electrodes remain connected to the battery during this process. Express your answer with the appropriate units. 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 1.3 cm apart? The electrodes remain connected to the battery during this process. Express your answer with the appropriate units.

Explanation / Answer

This is a problem of capacitors

Part a)

The expression for the capacitance is

C = Q/V = oA/d (1)

Data

V = 14 V

d = 0.50 cm= 0.50 10-2 m

diameter = = 13.0 cm = 13.0 10-2 m

I'm surprised you talk a little diameter parallel plate electrodes, the most common is the length of the plate, whatever the case with this parameter must calculate the area. I will calculate the area with the diameter

A = (d/2)2

A= ( 0.130/2)2

A= 1.327 10-2 m²

Calculate charge with 1

Q = oV A/d

Q = 8.854 10-12 14 1.327 10-2 / 0.5 10-2

Q = 3.29 10-10 C

Part b)

The electric field is

V = E d

E = V /d

E = 14/ 0.50 10-2

E = 28 102 N/C

Part c)

The potential difference is caused by the battery V= 14 V

Part d)

In part to find the expression for charge

Q = oV A/d

We calculate the new distance

Q = 8.854 10-1214 1.327 10-2/ 1.3 10-2

Q = 1.26 10-10 C

Part e)

The electric field is

V = E d

E = V /d

E = 14/ 1.3 10-2

E = 18 102 N/C

Part f)

The potential difference does not depend on the distance

V= 14 V

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