Two Questions on PHY 114 , Your answer will be rated if it is correct , Please d
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Two Questions on PHY 114 , Your answer will be rated if it is correct , Please don't post your answer, send it to my email calculusmind@gmail.com
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Please show every step.
1) Two electrons are initially 4.5 nm apart.
a) What is the magnitude of the electric force on either electron?
b)What will the force between the two charges be when the distance between them is increased to five times its original value?
2) Two charges, q1 = 5.8 ?C and q2 = 2.7 ?C are a distance 25.0 cm apart.
a) Calculate the magnitude of the electric field at the position of q2 due to q1,
b) the magnitude of the electric field at the position of q1 due to q2,
c) the magnitude of the electric force on either of the two charges.
Explanation / Answer
Charge of electrons Q1=Q2 =1.6x10^-19 C
separation
R = 4.5 nm = 4.5x10^-9 m
a)magnitude of the electric force
F =kQ1Q2/R^2 = 9x10^9*1.60x10^-19*1.60x10^-19/(4.5x10^-9)^2 = 1.13x10^-11 N
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b)
if the distance between them is increased by 5 times
F is inversely proportional to R^2
so decreased by 25 times
new force F' =1.13x10^-11/25 = 4.52x10^-13 N
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2)
Given
Q1 = 5.8x10^-6 C ,Q2 = 2.7 x10^-6 C
distance d=25 cm =0.25 m
a)
the magnitude of the electric field at the position of q2 due to q1
E1 = KQ1/d^2 = 9x10^9*5.8x10^-6/0.25)^2 = 8.35x10^5 N/c
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b)
the magnitude of the electric field at the position of q1 due to q2,
E2=KQ2/d^2 = 9x10^9*2.7x10^-6/(0.25)^2 = 3.88x10^5 N/c
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c)
Force = kQ1Q2/d^2 = 2.2545 N
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