Write the Coulomb’s law.
F=k|q1||q2|r2 (I)
Here, F is the electric force, k is a constant q1,q2 are the charges and r is the distance between the charges
The force due to the 0.80 μC charge is upward and that due to the 1.0 μC is to the right.
Write the equation for the magnitude of the net force on −0.60 μC .
F=Fx2+Fy2 (II)
Here, F is the magnitude of the net force on −0.60 μC , Fx is the magnitude of force on −0.60 μC due to 1.0 μC and Fy is the magnitude of force on −0.60 μC due to 0.80 μC
Write the equation for the direction of the net force on −0.60 μC .
θ=tan−1FyFx (III)
Here, θ is the angle the net force on −0.60 μC makes with the horizontal
Conclusion:
The value of k is 8.988×109 N⋅m2/C2 .
Refer to the figure and find the force on −0.60 μC due to 0.80 μC using equation (I).
Fy=(8.988×109 N⋅m2/C2)(0.80 μC(1 C106 μC ))(0.60 μC(1 C106 μC ))(8.0 cm(1 m100 cm))2=(8.988×109 N⋅m2/C2)(0.80×10−6 C)(0.60×10−6 C)(0.080 m)2=0.67 N
Refer to the figure and find the force on −0.60 μC due to 1.0 μC using equation (I).
Fx=(8.988×109 N⋅m2/C2)(1.0 μC(1 C106 μC ))(0.60 μC(1 C106 μC ))(10.0 cm(1 m100 cm))2−(8.0 cm(1 m100 cm))2=(8.988×109 N⋅m2/C2)(1.0×10−6 C)(0.60×10−6 C)(0.100 m)2−(0.080 m)2=1.5 N
Substitute 0.67 N for Fx and 1.5 N for Fy in equation (II) to find F .
F=(0.67 N)2+(1.5 N)2=1.6 N
Substitute 0.67 N for Fx and 1.5 N for Fy in equation (III) to find θ .
θ=tan−11.5 N0.67 N=24°
Therefore, the electric force on the −0.60 μC charge due to the other two charges is 1.6 N at 24° above the positive x-axis .