pos Dr. Zhou managed to put charge q = -64 mu C on in isolated electrically char
ID: 1338258 • Letter: P
Question
pos Dr. Zhou managed to put charge q = -64 mu C on in isolated electrically charged object, in his PHY II Lab. Which of the following statement is true is true for the Object which was initially neutral The object now: has additional 4 Times 10^14 electrons. lacks 4 Times 10^14 electrons has additional 3 Times 10^14 electrons. lacks 3 Times 10^14 electrons A 2.0 mu F Capacitor is fully charged using a 1.5 V battery. The charge on capacitor plate is: A 1.5 V battery is connected to a 5 Ohm resistor. The current through the resistor is: You are discharging A 2.00 mu F capacitor which is fully charged with voltage difference between the plates 12V (i.e., V_0=12V). A 5.0 M Ohm resistor is in the circuit in the series with capacitor. How long it will take to drop voltage between capacitor plates to 8.8 V?Explanation / Answer
MC1 : Addition of negative charge will add additional electrons to the object and -64 microC charge corresponds to
~4*1014 number of electrons. answer is option (a).
MC2 : charge on capacitor is given by formula Q = C*V
therefore Q = 2(microF) * 1.5 V which gives, Q = 3 microCoulombs answer is option (c).
MC3 : Using Ohm's law V=I*R therefore we get i =V/R
therefore, I = 1.5/5 = 0.3 Amperes answer is option (c).
MC4 : When capacitor charged to some voltage is connected to a resistor in series connection , the voltage across the capacitor is given by formula
Vc = Vo[exp(-t/RC)] , where Vo = initial voltage till capacitor is charged to
R = 5.0 Mohm C = 2.0 microFarad and t = time in seconds and
Vc = voltage across capacitor and R*C = 10 s-1
therefore Vc = 12 [exp(-t/10)] = 8.8 V which gives on solving
t = 3.1 seconds therefore answer is option (a).
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