Help with solutions! Help please? Thanks

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1 Answer
Jan 22, 2018

#"Details have shown below but please check my math."#

Explanation:

  • You firstly must draw a diagram to solve problem.

enter image source here The magnitude of force is the same as the values are the same.

  • Electric charges are on the corners of the isosceles triangle.
  • You must perform the necessary unit conversions.

#"50 cm=0.5 m"#
#+3 mu C=+3*10^-6C#
#+5 mu C=+5*10^-6C#
#-5 mu C=-5*10^-6C#

#sin alpha=40/50=4/5#
#cos alpha=30/50=3/5#

  • We know that electrical charges force each other.

#F=k*(q_1*q_2)/r^2#

  • Pushing force occurs between the electric charge at point A and the electric charge at point B.

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#color(blue)(F_(BA)=k*(q_A*q_B)/(0.5)^2=(9*10^9*5*10^-6*3*10^-6)/(0.25)=(135*10^-3)/0.25=540*10^-3=0.54N)#

  • A pulling force occurs between the electric charge at point A and the electric charge at point C.

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  • The magnitude of force is the same as the values are the same.

#color(green)(F_(CA)=-0.54N)#

  • We must find the vector sum of the blue and green forces.

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#F=sqrt(F_(BA)^2+F_(CA)^2+2*F_(BA)*F_(CA)*cos(2 alpha)#

#cos(2alpha)=cos^2 alpha-sin^2 alpha#

#cos(2alpha)=(3/5)^2-(4/5)^2=9/25-16/25=-7/25#

#F=sqrt((0.54)^2+(0.54)^2+2*0.54*(0.54)*(-7/25))#

#F=sqrt(0.2916+0.2916-2*0.2916*7/25)N#

#F=sqrt(0.5832-0.163296)#

#F=0.65 N#