What is the electric field at a distance of 2.0 m from a point charge of 40 J/C?

AI Thread Summary
The discussion revolves around calculating the electric field at a distance of 2.0 m from a point charge, given that the electric field at 1.0 m is 40 J/C. The correct formula for electric field strength is charge divided by the distance squared, leading to a calculation of 10 J/C at 2.0 m. Participants clarify that the correct unit for electric field strength is Newtons per Coulomb (N/C), not Joules per Coulomb (J/C). There is also a mention of using the constant K in the electric field equation to find the charge and subsequently the electric field at the specified distance. The conversation emphasizes the importance of using the correct units and formulas in physics calculations.
wakejosh
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Homework Statement



at a distance of 1.0 m from a point charge, the electric field is 40 J/C. What is the electric field at a distance of 2.0 m from the point charge?


The Attempt at a Solution



so, i understand that the elecric field should be = charge/distance(squared).

so I did:
40/(2^2) = 10 J/C

Is this correct or am i missing something? thanks
 
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Are you sure it's J/c? Isn't that volts? Shouldn't it be N/c?

http://www.glenbrook.k12.il.us/GBSSCI/PHYS/CLASS/estatics/u8l4b3.gif

The standard metric units on electric field strength arise from its definition. Since electric field is defined as a force per charge, its units would be force units divided by charge units. In this case, the standard metric units are Newton/Coulomb or N/C.

Excuse me if I'm blabbering, I'm just trying to look smart with my less than a year's worth of physics:biggrin:
 
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im no craque at physics, but all my options are in J/C.
 
E=\frac{Kq}{r^2}
k=\frac{1}{4\pi \epsilon}=9x10^9N
Find q, and then use the above equations to find E at r=2.
 
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