PHYS 142 Mid Tem Exam Spring Semester 2018 Name nstrections: For problems where
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PHYS 142 Mid Tem Exam Spring Semester 2018 Name nstrections: For problems where you need to write in answers, correet anits (S system) must accompany the units. Leaving off the units will cost you 1/3 rd the points for that answer, if the number is correct. alur or 1) (9 points) What is the magnitude of the electric force of attraction between a silver nuc hear (qr +47e) and its electrom the distance of 1.8x101°m? G-I- Dete ctric line carryir potential differe 2) (15 points). Calculate the electric potential at point A due to the two point charges Q1--75HC, and Q2-+75HC 52 Cm meter of: etic field 3) (8 points). How strong is the electric field between two plates 4.3 mm apart if the potential difference between them is 110 V? n must the nlates have if they are to be separated by a 1.2 mmaiExplanation / Answer
1)Ans
From the quantization of charge, the charge of the nucleus is given by,
q=47e
Here, e is the charge on one proton.
Substitute 1.6x10-19 C for e in the above expression.
q=47x(1.6x10-19)
=7.52x10-18C
The nucleus of an silver atom consists of protons and neutrons.
The charge on neutrons is zero so that they do not contribute in the net charge of the nucleus.
Only the protons contribute in the net charge of the nucleus.
The number of protons in the silver nucleus is 26 such that the net charge of the nucleus is 26 times the charge of a single proton.
F=(kq1q2)/r2
q1=charge of the nucleus =7.52x10-18 C
q2=charge of electron =1.6x10-19 C
k=9x109 Nm2/C2
given r=1.8x10-10m
F=(1.08x10-26)/(1.8x10-10)2
F=3.342x10-7 N
2.ans)
Given that Q1=-75C ,Q2=+75C
r1=30cm ,r2=52cm
VA=ke[(Q1/r1)+(Q2/r2)]
=(9x109)[(-2.5x10-6)+(1.4x10-6)]
=-9900V
3.ans)
E=V/d
Given V=110V,d=4.3mm
E=110/(4.3x10-3)
=2.5x104 V/m
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