Answer to Question #243509 in Electricity and Magnetism for wew

Question #243509

Consider three point charges at the corners of a triangle, as shown in the figure, where q1 = 8.00 x10^-9C, q2 = −4.00 × 10^−9 C, and q3 = 6.00 × 10^−9 C. Find the magnitude and direction of the resultant force on q3


1
Expert's answer
2021-09-28T16:55:15-0400

F23=kq2q342=9×109×4×6×101816F23=13.5×109  NF23x=13.5×109  NF23y=0F13=kq1q252=9×109×8×6×101825F13=17.28×109  NF13x=F13×cos(37)=17.28×109×0.7986=13.799×109  NF13y=F13×sin(37)=17.28×109×0.6018=10.399×109  NF_{23} = \frac{kq_2q_3}{4^2} = \frac{9 \times 10^9 \times 4 \times 6 \times 10^{-18}}{16} \\ F_{23} = -13.5 \times 10^{-9} \;N \\ F_{23x}= -13.5 \times 10^{-9} \;N \\ F_{23y}= 0 \\ F_{13} = \frac{kq_1q_2 }{5^2} = \frac{9 \times 10^9 \times 8 \times 6 \times 10^{-18}}{25} \\ F_{13} = 17.28 \times 10^{-9} \;N \\ F_{13x} = F_{13} \times cos(37) = 17.28 \times 10^{-9} \times 0.7986 = 13.799 \times 10^{-9} \;N \\ F_{13y} = F_{13} \times sin (37) = 17.28 \times 10^{-9} \times 0.6018 = 10.399 \times 10^{-9} \;N

Total force:

Fx=F23x+F13x=13.5×109+13.799×109=0.299×109  NFy=F23y+F13y=0+10.399×109=10.3399×109  NMagnitude=Fx2+Fy2=1.0344×108  NF_x=F_{23x}+F_{13x}=-13.5 \times 10^{-9} + 13.799 \times 10^{-9} = 0.299 \times 10^{-9} \; N \\ F_y=F_{23y}+F_{13y}= 0 + 10.399 \times 10^{-9} = 10.3399 \times 10^{-9} \; N \\ Magnitude= \sqrt{F_x^2+F_y^2}= 1.0344 \times 10^{-8} \;N

angle with +ve x axis =arctan(FyFx)=72.07  degrees=arctan(\frac{F_y}{F_x})= 72.07 \; degrees


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment