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An electron is placed at point P in the electric field set up by a source charge, Q. Point P is located 50 cm from Q and has an electric field strength of 1.08x10^5 N/C directed away from Q. What is the magnitude of charge Q?
da_willem
Mar17-04, 01:42 AM
Coulombs law:
E(r) = \frac{Q}{4 \pi \epsilon r^2}
gives the electric field strength E(r) at a distance r from a point charge Q. The electrical permittivity \epsilon can be found in a tablebook. For vacuum or air it is approx. 8.85e-12 F/m.
Chi Meson
Mar17-04, 11:00 AM
Just a comment:
the fraction \frac{1}{4 \pi \epsilon}
is equal to the coulomb constant "k" (8.99 E9)
Originally posted by Chi Meson
Just a comment:
the fraction \frac{1}{4 \pi \epsilon}
is equal to the coulomb constant "k" (8.99 E9)
Only when:
\epsilon = \epsilon_0
If you are in a different medium there is a relative permeability \epsilon_r in which case:
\epsilon = \epsilon_0\epsilon_r
And K is different.
Chi Meson
Mar17-04, 12:43 PM
Agreed.
It was my assumption that the person asking the question was not yet at that level, and might have been taken aback by the use of epsilon when the textbook uses "k."
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