anonymous
  • anonymous
An electric field w/ a magnitude of 154 N/C exists @ a spot .10 m away from a charge. @ a place that is .33 m away from this charge, what is the electric field strength?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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JamesJ
  • JamesJ
Electric field varies as 1/r^2, where r is the distance from the charge. So scale the field for a change of r from 0.10 m to 0.33 m
anonymous
  • anonymous
can you do this step by step please?
JamesJ
  • JamesJ
The electric field strength of a point charge Q at a distance r from Q is given by \[ E = \frac{kQ}{r^2} \] where k is a positive constant. We know that when r = 0.1 m that E = 154. That is \[ 154 = \frac{kQ}{0.1^2} \ \implies kQ = 154(0.01) = 1.54 \] Now use that to find the electric field, E, when r = 0.33. That is, use \[ E = \frac{kQ}{r^2} \] You now know kQ and r for this new situation.

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JamesJ
  • JamesJ
Make sense?

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