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?
a)E=kq/r2,k=9×109a) E=kq/r^2, k=9\times 10^9a)E=kq/r2,k=9×109
E=−(9×109×3×10−9)0.252=−E=-\frac{(9\times 10^9 \times3\times 10^-9)}{0.25^2}=-E=−0.252(9×109×3×10−9)=− 432N/C432N/C432N/C
b) 12=9×109×3×10−9r212=\frac{9\times 10^9 \times 3\times 10^-9}{r^2}12=r29×109×3×10−9
r2=27/12=2.25r^2=27/12=2.25r2=27/12=2.25
r=1.5mr=1.5mr=1.5m
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment