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Q # 1
A parallel plate air capacitor of capacitance 245pF has a charge of magnitude 0.148μC on each plate. The plates are 0.328mm apart.
(a) What is the potential difference between the plates?
(b) What is the area of each plate?
(c) What is the electric field magnitude between the plates?
(d) What is the surface charge density on each plate? Marks 8
Q # 2
A cylindrical copper cable 1.50Km long is connected across a 220.0V potential difference.
(a) What should be its diameter so that it produces heat at a rate of 50.0W?
(b) What is the electric field inside the cable under these conditions? Marks 6
Q # 3
A person with body resistance between his hands of 10KΩ accidently grasps the terminal of a 14-KV power supply.
(a) If the internal resistance of the power supply is 2000Ω, what is the current through the person’s body?
(b) What is the power dissipated in his body?
(c) If the power supply is to be made safe by increasing its internal resistance, what should the internal resistance be for the maximum current in the above situation to be 1.00mA or less? Marks 6
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Q#2:
First, the value of the length of the cable will be converted to meters.
Therefore, the length of the cable is L = 1.5*10^3 m = 1500 m.
We know, from enunciation, the value of potential difference:V = 220 volts
The power is given and it's value is of P = 50 Watts
P = V^2/R => R = V^2/P
Since the area of the section of the cable is circular, w'ell recall the formula for the area of the circle:
A = pi*r^2 (1)
A = p*L/R (2), where p = 1.72/10^8 ohm/m
We'll equate (1) and (2) and we'll get:
pi*r^2 = p*L*P/V^2
r^2 = p*L*P/V^2*pi
r = sqrt(p*L*P/V^2*pi)
Diameter is d = 2*r = 2*sqrt(p*L*P/V^2*pi)
d = 2*sqrt1.72*10^-10*15*5*10^3/484
d = 2*sqrt 1.72*10^-7*75/22
d = 1.0325/100
d = 0.010325 meters
The diameter of the cable is of d = 0.010325 meters, such as it is producing heat at a rate of 50W.
Q#1:
(a)we have q=cv where q=0.148μC
(remember we only put the magnitude of
one plate and we don't double it) and
c=245pF, so
v(potential difference)=604.08 volt
(b)c=(8.85*10^-12)*A/d where d=0.328mm,
so A(area)=9.08*10^-3 .
(c)v=Ed so E=v/d,
E(electric field)=1.84*10^6 v/m
(d)surface charge density=q/A=1.63*10^-5
PHY101 Assignment#04 Complete Solution Spring 2011
See the attached file please
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