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

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SUMMARY

The electric field at a distance of 2.0 m from a point charge of 40 J/C is calculated using the formula E = k * q / r², where k is Coulomb's constant (approximately 9 x 10⁹ N m²/C²). The initial calculation of 10 J/C is incorrect; the correct unit for electric field strength is Newtons per Coulomb (N/C), not Joules per Coulomb (J/C). The discussion clarifies that the electric field strength decreases with the square of the distance from the charge, confirming that at 2.0 m, the electric field is indeed 10 N/C.

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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:
 
Last edited by a moderator:
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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