Two 10-cm-diameter electrodes 0.52 cm apart form a parallel-plate capacitor. The

Question

Two 10-cm-diameter electrodes 0.52 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 20 Vbattery.

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What are a) the charge on each electrode, b) the electric field strength inside the capacitor, and c) the potential difference between the electrodes while the capacitor is attached to the battery?

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What are a) the charge on each electrode, b) the electric field strength inside the capacitor, and c) the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.0 cm apart? The electrodes remain connected to the battery during this process.

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What are a) the charge on each electrode, b) the electric field strength inside the capacitor, and c) the potential difference between the electrodes after the original electrodes (not the modified electrodes of parts D-F) are expanded until they are 20 cm in diameter while remaining connected to the battery?

Explanation / Answer

Expression for capacitance:

C = *A/d

Here, area A = (r)2 = (0.10 m/2)2 = 7.85 x 10-3 m2

Distance: d = 0 .52/100 = 5.2 x 10-3   m

Now sunstitute these values in   C = *A/d

C = ( 8.854*10-12 ) (7.85 x 10-3 m2 )  / 5.2 x 10-3   m

= 13.36 x 10 -12 F= 13.36 pF

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Now find charge:

Q = ( C ) ( V )

= ( 13.36 x 10 -12 F ) (20 )

= 267.3 x 10 -12 C  

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b)

The electric field is the voltage divided by the separation distance:

E = V/d

= 20 V / ( 5.2 x 10-3   m)

= 3.85 x 103 V/m

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c)

The potential difference is northing but the battery voltage, since the voltage will remain constant immediately after battery disconnection:
V = 20 V

( please post other questions separately)

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