Answer to Question #194545 in Electric Circuits for Plea

Question #194545

A particle has charge -3 nC. A) Find the magnitude and direction of the electric field due to this particle at a point 0.25 m directly above it. B) At what distance from this particle does its electric field have a magnitude of 12 N/C?


1
Expert's answer
2021-05-19T20:41:15-0400

a)E=kq/r2,k=9×109a) E=kq/r^2, k=9\times 10^9

E=(9×109×3×109)0.252=E=-\frac{(9\times 10^9 \times3\times 10^-9)}{0.25^2}=- 432N/C432N/C


b) 12=9×109×3×109r212=\frac{9\times 10^9 \times 3\times 10^-9}{r^2}

r2=27/12=2.25r^2=27/12=2.25

r=1.5mr=1.5m


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