For problem 12, start from a fundamental principle and be sure to clearly show a

Question

For problem 12, start from a fundamental principle and be sure to clearly show
all work|your score will be based on your reasoning. Include units and use 3-component
vector notation when appropriate. Additional paper will be provided if needed..

Problem 12:
An uncharged capacitor is connected to a power supply, in series with a resistor, as shown in the diagram, and current begins to run in the circuit. The square plates of the capacitor have dimensions 2m x 2m, and are separated by a gap of 0.5 mm, filled with air. The resistor has resistance 300 ohms, length 10 cm, and radius 5 mm. The emf of the power supply is 2000 volts.

A) When the electric field in the capacitor gap increases beyond 3x106 N/C, a spark forms in the gap, and
the air becomes a conductor. At the instant just as before the spark begins to form, what is the magnitude
of the potential difference across the capacitor?
B) At the instant just before you see the spark, how much charge is on the positive plate of the capacitor?
C) At the instant just before you see the spark, how much current is running through the resistor?

Problem 12(30 pts: An uncharged capacitor is connected to a power supply, in series with a resstor, as shown in the diagram. and current begins to run in the drait. The square platesof the capacitor have dimensions 2m 2m, and are separated by a gap of 0.5 mm, Oled with air. The resisor has resistance 300 0, ength 10 cm, and radius 5 mm. The emf of the power supply is 2000 volts A) When the electric oeld in the capacitor gap inaeases beyond 3 10 NC, a spark forms in the gap, and the air becomes aconductor. At theinstant just as before the spark beginsto form, what is the magnitude of the potential dioerence across the capacitor? B) At the instant just before you see the spark, how much charge is on the positive plate of the capacitor? c) At the instant just before you see the spark, how much current is running through the r

Explanation / Answer

a) E = 3e6 N/C

V = Ed

= 3e6*0.5e-3

= 1500 V answer

b) Charge = CV = VAeo/d

= 1500*2*2*8.85e-12/0.5e-3

= 0.0001062 C

c) current Vr/R = (2000-1500)/300 = 1.667 A

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