1.The net charge on the planet earth is equivalent to the charge on the number o

Question

1.The net charge on the planet earth is equivalent to the charge on the number of electrons contained in one US penny ( pre- 1982). Assume this charge is distributed evenly over the surface of the earth . A) Determine the area charge density on the surface of the earth. B) Determine the strength of the electric field at a point ten meters above the surface of the the earth due to this charge. C) A typical lightning bolt transfers approximately 10 C to 50 C of charge . Discuss this fact conceptually in light of the net charge on the planet earth.

2. A nucleus of the most common isotop of the element with atomic number 11 is accelerated from rest through a potential difference of 2 kv. It then encounter a uniform E- field ( directed opposite the direction of its velocity) coming to a complete stop in a distance of 93500 A. Determine the strength of the E-field .Also, determine the average current ( whether measurable or not) due to the presence of the nucleus as it moved in the E- field ( Justify the answer)

Explanation / Answer

1.

a)

Mass of US penny

M=2.5 g (from wikipedia)

Atomic number of copper Z=29

Atomic weight of copper =63.546 g/mol

Number of electrons in penny

n=NA*No of moles*Atomic number =(6.02*1023)(2.5/63.546)*29

n=6.868*1023 electrons

Charge of electron q=1.602*10-19C

Total charge

Q=nq =(6.686*1023)(1.602*10-19)

Q=1.1003*105 C

Radius of earth

R=6371 Km

Surface area

A=4piR2=4pi*(6371*103)2 =5.1*1014 m2

Charge density

sigma =Q/A =1.1003*105 /5.1*1014

sigma =2.157*10-10 C/m2

b)

Electric field

E=KQ/r2

here r=6371*103+10 =6371*103 km (approx)

E=(9*109)(1.1&105)/(6371*103)2

E=24.4 N/C

c)

Compared to net charge of earth ,the lighting bolt has a charge of very small magnitude .

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