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    The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C....

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atrinity | Honors

Posted May 2, 2013 at 6:16 PM via web

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    The strength of an electric field at 5.0 cm from a point charge is 100.0 N/C. What is the magnitude of the source charge? Show your work. 

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ishpiro | Teacher | (Level 1) Associate Educator

Posted May 3, 2013 at 1:48 AM (Answer #1)

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`` The electric field at the distance d away from the point charge is given by

`E = k q/d^2`

where `k` is the constant: `k = 8.99*10^9 (N*m^2)/C^2`

Thus, the magnitude of the point charge is

`q = (E*d^2)/k`

Plug in the values for E and d, making sure to convert d = 5 cm into meters:  d = 0.05 m.

`q = ((100 N/C) *(0.05 m)^2)/(9*10^9 (N*m^2)/C^2)=2.8*10^(-11) C`

The magnitude of the charge is `q=2.8*10^-11 C`

 

 

 

 

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