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A point charge of $5\ \mu$C is placed at the origin. The electric field magnitude at $0.5$ m is:
A$1.8\times 10^5$ N/C, by $E = kQ/r^2$
B$9\times 10^4$ N/C, half the correct value here
C$0$ N/C, since the point is far on the table
D$5\times 10^{-6}$ N/C, the charge in coulombs
Answer & Solution
Correct answer: A. $1.8\times 10^5$ N/C, by $E = kQ/r^2$
$E = 9\times 10^9\cdot 5\times 10^{-6}/0.25 = 1.8\times 10^5$ N/C.
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