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Physics
OpenStudy (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?

OpenStudy (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

OpenStudy (anonymous):

can you do this step by step please?

OpenStudy (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.

OpenStudy (jamesj):

Make sense?

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