HCV2 Page no.121 22

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HCV2 Page no.121 22

Post  gaurav on Sat Jun 07, 2008 9:23 am

HCV2 Page no.121 22

gaurav

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Re: HCV2 Page no.121 22

Post  Pinnaka on Sun Jun 08, 2008 4:25 am

a. the electrostatic force can be calculated easily by using coulombís law. The distance between the charges is given to be 0.050m
using
F-el = k-e*q^2/r^2 = 0.144N


c. The tension in the string has to balance out the net force (resultant ) force of electrostatic force and weight

T^2 = mg^2 + F-el ^2
T turns out to be 0.986 N

d, b

Hmm, interesting problem

First look at the conditions given it doesnít say they (charges) are in electrostatic equilibrium. I will skip some of the details. When there is such a configuration and the charges are said to be in equilibrium then the following expression follows

mg tan (theta) = f-el
f-el is different from 0.144N, The charges under repulsive electrostatic force tend to moves away from each other .now let us examine if the charges are in electrostatic equilibrium or not?
sin (theta) = 0.025/0.5= 0.05
so now cos (theta ) can also be calculated. From the value of sin (theta) we can say that theta is very small and the string is almost vertical
Tcos(theta) = mg
T turns out to be 0.098N (equilibrium along y axis)
then Tsin (theta) the component of force is equal to 0.098*0.05 = 4.90*10^-3
the net force along the x axis is 0.144 ĖTsin (theta) = 0.0948N
The force is away from the charges so the net force along the y axis is zero and the x axis is given above

d. acceleration is F/m = 0.095m/s^2

Ps: gaurav thanks for posting the problem. But however u should post the entire text of the problem because the experts who take the question are very busy grad and undergrad students (research). Since this is u r first post it has been answered. Do not repeat this in the future.

Hope this Helps
Pinnaka

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