Two 10-cm-diameter electrodes 0.46 cm apart form a parallel-plate capacitor. The

Question

Two 10-cm-diameter electrodes 0.46 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 17 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged.

(a) What is the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes right after the battery is disconnected?

(b) What are the charge on each electrode, the electric field strength inside the capacitor, and 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?

(c) What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes after the original electrodes (not the modified electrodes of part (b)) are expanded until they are 20 cm in diameter?

Explanation / Answer

a) q1 = C*v ..............(1)
where C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.05)^2/0.0057
so q1 = -q2, can be calculated by equation 1

b) E = /o = q/(8.854x10^-12**r^2) =

c) V=E(V/m)*0.0046 = # E from part b

d) Now C decreases to
C = o*A/d = 8.854x10^-12**r^2/d = 8.854x10^-12**(0.05)^2/0.01 =

q1 = -q2 Charge remains constant in this situation

e) E = E(V/m )It remains the same as in part b

f) V = E*d = E *0.01=

keep all values

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