Question

Help with solutions! Help please? Thanks

Answer 1

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

Explanation:

• You firstly must draw a diagram to solve problem.

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.

`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.

• 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.

`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`