Calculating Electric Field Strength and Direction for a Negative Charge

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dragon-kazooie
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I think I have this correct, could someone please verify? I compared answers with a friend who got 2.0 x 106 N/C to the left, and I can't figure out how they got that answer, so I'm hoping mine is right...

1. Homework Statement

A negative charge of 3.5 × 10–8 C experiences a force of 0.070 N to the right in an electric field. What is the field magnitude and direction?

Homework Equations


Electric field strength = f / q

The Attempt at a Solution



Electric field strength = f / q

Electric field strength = 0.070 N / 3.5x10-8

Electric field strength = 2x106 N/C To the left
 
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Funny, ##2\times 10^6## N/C to the left looks to me to be the same as ##2\times 10^6## N/C to the left . Or do you mean you found ##2\times 106 = 212 ## N/C to the left ?
 
BvU said:
Funny, ##2\times 10^6## N/C to the left looks to me to be the same as ##2\times 10^6## N/C to the left . Or do you mean you found ##2\times 106 = 212 ## N/C to the left ?

Oh goodness, I got myself all mixed up. I should have written "I compared answers with a friend who got 0.02 N/C to the left, and I can't figure out how they got that answer, so I'm hoping mine is right..."

Thank you for asking for clarification. I'm sorry that I didn't proof read my question closely enough!
 
dragon-kazooie said:
Electric field strength = 2x106 N/C To the left
Once you fix your exponent, this looks correct to me.
dragon-kazooie said:
a friend who got 0.02 N/C
Not sure how your friend could have gotten this. Maybe ask them to show your their work?
 
Thank you!