Question

A point charge of +18 μC is on the y axis at y = +3.00 m.A point charge of -12 μC is at the origin.A point charge of +45 μC is on the x axis at x = +3.00 m. Find the angle, in degrees, of the force relative to the negative x-axis.

Answer 1

The magnitude of the resultant force is 0.383 N.

The angle of the force is 48.6⁰.

How to calculate the magnitude and angle of the force?

The force in x - direction is calculated as;

Fx = kq₁q₂/r²

Fx = (9 x 10⁹ x 45 x 10⁻⁶ x 12 x 10⁻⁶ ) / ( 3²)

Fx = 0.54 N

The distance between the two charges;

r = √ (3² + 3²)

r = 4.24 m

The force between the two charges along x -axis;

Fx = -(9 x 10⁹ x 45 x 10⁻⁶ x 18 x 10⁻⁶ ) / ( 4.24²) x cos(θ)

Fx = -(9 x 10⁹ x 45 x 10⁻⁶ x 18 x 10⁻⁶ ) / ( 4.24²) x (3 / 4.24)

Fx = -0.287 N

The force between the two charges along y - axis;

Fy = (9 x 10⁹ x 45 x 10⁻⁶ x 18 x 10⁻⁶ ) / ( 4.24²) x sin(θ)

Fy = (9 x 10⁹ x 45 x 10⁻⁶ x 18 x 10⁻⁶ ) / ( 4.24²) x (3 / 4.24)

Fy = 0.287 N

The total force in x direction;

x = 0.54 N - 0.287 N

x = 0.253 N

The total force in y - direction

y = 0.287 N

The magnitude of the resultant force is;

F = √ (0.253² + 0.287²)

F = 0.383 N

The angle of the force is calculated as follows;

θ = arc tan (y/x)

θ = arc tan (0.287 / 0.253)

θ = 48.6⁰

Answer 2

Using Coulomb's law:

`\begin{gathered} F_(21)=K(Q1\cdot Q2)/(r^2) \\ F_(21)=8.988*10^9\cdot(45*10^(-6)\cdot(-12*10^(-6)))/(3^2) \\ F_(21)=0.53928N \end{gathered}`
`\begin{gathered} F_(31)=K(Q1\cdot Q3)/(r^2) \\ F_(31)=8.988*10^9\cdot\frac{45*10^(-6)\cdot(18*10^(-6))}{(\sqrt[]{18})^2^{}} \\ F_(31)=0.40446N \end{gathered}`
`\begin{gathered} F_(Tx)=F_(T21)\cos (45) \\ F_(Ty)=F_(T31)\sin (45) \end{gathered}`
`\begin{gathered} |F_T|=\sqrt[]{(F_x)^2+(F_y)^2}= \\ |F_T|=\sqrt[]{(-0.25328)^2+(0.285996)^2} \\ |F_T|\approx0.382 \end{gathered}`

And the angle:

`\begin{gathered} \theta=\tan ^(-1)((F_y)/(F_x)) \\ \theta\approx-41.52868^{} \end{gathered}`