A positively charged particle Q1 = +25 nC is held at fixed position. A second charge Q2 of mass m = 6.5 μg is located a distance d = 25 cm directly above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth. Calculate the magnitude of Q2 in units of nanocoulombs.  This is the equation right? --> |Q2| = ( m g d2 )/(

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Chapter1: Units, Trigonometry. And Vectors
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 A positively charged particle Q1 = +25 nC is held at fixed position. A second charge Q2 of mass m = 6.5 μg is located a distance d = 25 cm directly above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.

Calculate the magnitude of Q2 in units of nanocoulombs. 

This is the equation right? --> |Q2| = ( m g d2 )/( k Q1 )  

I got 1.76 but was told that my answer was off by by a factor of 10n. Not sure where to go from here. 

Q2
d
Q,
Transcribed Image Text:Q2 d Q,
Expert Solution
Step 1: Determine the given data:


straight Q subscript 1 equals space 25 space nC equals 25 cross times 10 to the power of negative 9 end exponent space straight C
Mass space of space straight Q subscript 2 equals 6.5 space μg equals 6.5 space cross times 10 to the power of negative 9 end exponent space kg
Distance space left parenthesis straight d right parenthesis equals 25 space cm equals 0.25 space straight m


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