Point charge A with a charge of +3.00 μC is located at the origin. Point charge B with a charge of +6.00 μC is located on the x axis at x = 7.00 cm. And point charge C with a charge of +2.00 μC is located on the y axis at y = 6.00 cm. What is the net force (magnitude and direction) exerted on each charge by the others?

charge A magnitude

charge A direction

charge B magnitude

charge B direction

charge C magnitude

charge C direction

A)  F_{A} = 36.26 N , θ = 204.4º

B) F_{B} = 43.45 N ,  θ = 349º

C)    F_{C} = 25.16 N ,  θ = 112.52º

Explanation:

The force is a vector quantity, so we must take into account the direction of the force, for the calculation of the electric force we will use Coulomb's law

         F₁₂ = k q₁ q₂ / r₁₂²

we will use numerical indexes 1 for A, 2 for B and 3 for C

a) Force on charge A

X axis

          Fₓ = F₁₂

          Fₓ = k q₁ q₂ / r₁₂²

          Fₓ = 8.99 10⁹ 3 10⁻⁶ 6 10⁻⁶ / 0.07 2

          Fₓ = 3,302 10¹ N

Since the two charges are of the same sign, the force is repulsive, so that it is directed towards the negative side of the x-axis.

Y Axis  

          = F₁₃

         F_{y} = k q₁q₃ / r₁₃²

         F_{y} = 8.99 10⁹ 3 10⁻⁶ 2 10⁻⁶ / 0.06 2

         F_{y} = 1,498 10¹ N

Repulsive force in the direction of the negative part of the y axis

To find the magnitude we use the Pythagorean Theorem

        = √ (Fₓ² + ²)

        F_{A}= √ (33.02 2 + 14.98 2)

        F_{A} = 36.26 N

For the direction let's use trigonometry

        tan θ = F_{y} / Fₓ

        θ = tan⁻¹ F_{y} / Fₓ

        θ = tan⁻¹ (-14.98 / -33.02)

        θ = 24.4º     in the third quadrant

Measured from the positive side of the x axis

       θ = 180 + 24.4

       θ = 204.4º

b) Force on charge B

     X axis

             Fₓ = F₂₁ + F₂₃ₓ

             F₂₁ = k q₁ q₂ / r₁₂²

             F₂₁ = 8.99 10⁹ 3 10⁺⁶ 6 10⁻⁶ / 0.07²

             F₂₁ = - F₁₂ = 33.02 N

Directed towards the positive side of the x axis

             F₂₃ = k q₃q₂ / r₃₂²

             R₃₂² = (0.07² +0.06²) = 0.0085

            F₂₃ = 8.99 10⁹ 2 10⁻⁶ 6 10⁻⁶ / 0.0085

            F₂₃ = 1,269 10¹ N

For the direction of this repulsive force we use trigonometry

          Tan θ = y / x

           θ = tan⁻¹ 0.06 / 0.07

           θ = 40.6º

As the force is repulsive therefore the angle is

            θ = 360 - 40.6

            θ = 319.4º

Now we can find the components of this force

            F₂₃ₓ = F₂₃ cos 319.4

            F₂₃ₓ = 12.69 cos 319.4

            F₂₃ₓ = 9.635 N

            = F₂₃ sin319.4

            F_{23y} = 12.69 sin319.4

            F_{23y} = -8.258 N

The force on this charge is

X axis

           Fₓ = F₂₁ + F₂₃ₓ

           Fₓ = 33.02 + 9.635

           Fₓ = 42,655 N

Y Axis

           F_{y} = F_{23y}

           F_{y} = -8.258 N

The Magnitude is

            = √ (Fₓ² + F_{y}²)

           F_{B} = √ (42,655 2 + 8,258 2)

           F_{B} = 43.45 N

The address is

             Tan θ = F_{y} / Fₓ

             θ = tan⁺¹ (-8.258 / 42.655)

             θ  = -10.96

Measured from the positive side of the x axis is

             θ = 360 -10.96

             θ = 349º

c) The force on the charge C

   X axis

          Fₓ = F₃₂ₓ

          Fₓ = -F₂₃ₓ = -9.635 N

Y Axis

         F_{y} = F₃₁ +

         F_{y} = 14.98 + 8.258

         F_{y}= 23,238 N

The magnitude is

              = √ (Fₓ² + F_{y}²)

              F_{C} = √ (9,635 2+ 23,238 2)

              F_{C} = 25.16 N

The address is

              tan θ = F_{y} / Fₓ

              θ = tan-1 (-23.238 / 9.635)

              θ = -67.48

Measured from the x axis

              θ = 180 -67.48

              θ = 112.52º

answer:

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