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Illuminitwit
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Homework Statement
A man walks into an electric field with a small charged object that has a charged value of 4.2 E-6C. At a distance of 10m from the object that is responsible for the electric field, he notices a force of 8.0 N on the object he is carrying.
What is the electric field intensity?
What is the potential difference?
Homework Equations
E = kc(q/r^2)
F = kc(q1q2/r^2)
∆V = -Ed
The Attempt at a Solution
F = kc(q1q2/r^2)
8.0 = 9 E9 • (4.2 E-6 • q2) / 10^2
800 = 9 E9 • 4.2 E-6 • q2
800/(9 E9 • 4.2 E-6) = q2 = 2.1 E-2
E = kc(q/r^2)
E = 9 E9 • (4.2 E-6 / 10^2)
E = 378
Round to two significant figures:
E = 380
[Interesting, but unnecessary; F/E = q2]
∆V = -Ed
∆V = -380 • 10
∆V = -3800
I think the first set of equations where I used F were unnecessary.
The electric field intensity = 380 N/C.
The potential difference is = -3,800 Volts.
I really don't think I did this right. The numbers I ended up with look too large, and I don't know if volts are supposed to be negative. I don't understand what "q" is supposed to be. Is it the difference between q1 and q2? What's q0? Does it even have a real value? I went through all the material and it still doesn't make sense!