Two 15-cm-diameter electrodes 0.58 cm apart form a parallel-plate capacitor. The

Question

Two 15-cm-diameter electrodes 0.58 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 17 Vbattery. After a long time, the capacitor is disconnected from the battery but is not discharged.

Part B:

What are the charge on each electrode (Q), the electric field strength inside the capacitor (E), and the potential difference (Delta V) between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.2 cm apart?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

Initial Capacitance, C = *A/d

A = *r^2 = *(.15/2)^2 = 1.77*10^-2 m^2
d = .58/100 = 5.8*10^-3 m

C = 8.854*10^-12 * 1.77*10^-2 / 5.8*10^-3
C = 27*10^-12
C = 27 pF

Q = C*V = 27*10^-12 * 17
Q = 4.59 * 10^-10 C

Charge on Each Electrode, Q = 4.59 * 10^-10 C

Now C decreases to
C = 8.854*10^-12 * 1.77*10^-2  / 0.012  
C = 13 pF

Q = C*V
V = Q/C
V = 4.59 * 10^-10 / 13 * 10^-12
V = 35.3 volt
Potential difference between the electrodes , V = 35.3 volt

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

E = V/d = 35.3/0.012
E = 2.94 * 10^3 V/m
Electric field strength inside the capacitor, E = 2.94 * 10^3 V/m

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