Oil Drop Electric Field Problem

AI Thread Summary
The discussion centers on calculating the charge of a negatively charged oil drop in an electric field of 6,400 N/C at a distance of 1.2 m. The relevant equation, E = kq/r², is acknowledged, but the user struggles with rearranging it to solve for charge (q). Attempts to isolate q using different methods yield incorrect results, prompting a reminder about the importance of units in calculations. Clarification is provided that k equals 9 x 10^9 N m²/C², emphasizing that proper unit handling is crucial for accurate answers. Ultimately, the user realizes the need to square root the product of k and r² to find the correct charge value.
yasemonkey
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Homework Statement


An oil drop is charged negatively. How much charge is on the drop if the Electric Field is 6,400 N/C at a distance of 1.2 m?

Homework Equations


E=k q/r^2
I don't know if I should be using that equation for this one.

The Attempt at a Solution


I tried to do E/(k x r^2) = q but that didn't work
I also tried E/r^2= q but that also didn't work

I know the answer has to be 1.3 x 10^3

Thanks in advance for all the help!
 
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You have the correct equation for field strength, you are just not solving correctly for q.

A good check is whether or not your resulting expression has the correct physical units.
 
hmm... okay

and I'm right in saying k= 9 x 10^9
 
No. As many other constants, k is dimensionful and k = 9 x 10^9 N m^2/C^2 - units are important and the units of your final answer must also make sense.
 
Orodruin said:
No. As many other constants, k is dimensionful and k = 9 x 10^9 N m^2/C^2 - units are important and the units of your final answer must also make sense.
Oh I have to square root k x r^2 !
 
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