Hai! Calculating Net Electrostatic Force! Need halp :(

  • Thread starter Mori
  • Start date
  • #1
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Homework Statement


Three charged objects are placed as shown. Find the net force on the object with the charge of -5uC
Diagram:[/B]
http://imgur.com/xHGHGbd

Homework Equations


No idea what to do as far as steps go, I don't know how to start :([/B]


The Attempt at a Solution


None so far.
 

Answers and Replies

  • #2
694
114
Try calculating the magnitude of the forces (individually) acting on the specified object and determine their nature (i.e. attractive/repulsive) , then find the resultant vector of force using trigonometry.
 
  • #3
8
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Okay, I'll go find each one acting on it, gimme a sec :D
 
  • #4
8
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6uc vs. -5uc = -0.0902135052958669(Attractive)
-7uc vs. -5uc = 0.07875 N(Repulsive)

How do I use these in trigonometry?
 
  • #5
694
114
You've made a unit conversion error somewhere - it should be 9.023...N and 7.875 N.
 
  • #6
8
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Oh, sorry xD
I might've typed it wrong,or I'm just kinda dumb.
 
  • #7
694
114
The forces will be vectors, so construct a vector triangle and find the "length of the missing side" in terms of Newtons.
 
  • #8
694
114
Not true - even PhD students can make conversion errors.
 
  • #9
8
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I got 34.6456i, 36.68 deg. S of E, did I get it? :D
 
  • #10
694
114
I'm afraid that is incorrect; how can the force have a magnitude which is non-real in this case? I suggest that you use the cosine rule to find the resultant force.
 
  • #11
8
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I did these, what did I do wrong?
Multiply it by sin(30) and multiply 7.875 again by cos(30) = (6.82 N, 3.94 N)
Turn the 6uc into vector components. = (-9.02 N,0)
Add the two vectors = (-2.2, 3.94)
Use pythagorean theorem on it = 3.27 N
Arctangent to find direction= -60.809688481761 degrees
3.37 N, 60 degrees N of W
 
  • #12
694
114
The vector triangle will not be right angled - do you have a justification for assuming one of the angles between the vectors to be 90 degrees? The resultant won't be on the same line as the 10cm one, so using pythagoras' principle won't work. Try using the cosine rule.
 
  • #13
8
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4.68 N, 57.4deg S of E
Got that after various tries, done?
 
  • #14
8
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Multiply it by sin(30) and multiply 7.88 again by cos(30) = (6.82 N, -3.94 N)
Turn the 6uc into vector components. = (-9.34 N,0)
Add the two vectors = (-2.52, 3.94)
Use pythagorean theorem on it = 4.68 N
Arctangent to find direction= -57.4 Degrees
4.68 N, 57.4deg S of E
 

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