Question

Four charges of equal magnitude, q = +1 uC, are placed at the four corners of a square with a side length of 1 meter located in the xy-plane. A small ball with a known charge +2 uC and an unknown mass is placed 2 meters above the center of the square (so along the z-axis).
Required:
a. What is the total force on the small ball from the four charges?
b. If the ball is in equilibrium where it was placed and we assume standard earth gravity, 9.8 m/s^2, what must the mass of the ball be?

Answer 1

Answer:

a)     F _total = 15.08 10⁻³ N ,   b)   m = 1,539 10⁻³ kg

Explanation:

a) To solve this problem we can use Coulomb's law to find the force created by each charge

          F =$k \frac{\ q_{1} q_{2} }{r_{12}^{2} }$

and the total force is

          F = F₁₅ + F₂₅ + F₃₅ + F₄₅

The bold are vectros, In the exercise they indicate the charges

          q₁ = q₂= q₃ = q₄ =  1 10⁻⁶ C

          q₅ = 2 10⁻⁶ C

Let's set a reference system where the sphere is on the z axis and the other charges on the xy plane, let's write the coordinates of card charge

sphere (subscript 5)

   x₅ = 0

   y₅ = 0

   z₅ = 2 m

charge 1

   x = 0.5 m

   y = 0.5 m

   z = 0

charge 2

   x = -0.5 m

   y = 0.5 m

   z = 0

 

charge 3

   x = -0.5 m

   y = -0.5 m

   z = 0

charge 4

   x = 0.5 m

   y = -0.5 m

   z = 0

With these values ​​we can calculate the distance between each charge and the sphere

charge 1 and sphere

       r₁₅² = (x₅ -x₁)² + (y₅ - y₁)² + (z₅ -z₁)²

substitute

       r₁₅² = (0- 0.5)² + (0 - 0.5)² + (2 -0)²

       r₁₅² = 0.5² + 0.5² + 2²

       r₁₅² = 4.5

we can see that for the other charges the result is the same since being squared always gives positive

      r₁₅ = r₂₅ = r₃₅ = r₄₅

the force created by the card charge on the sphere is the projection on the Z axis of the total force

Let's find the angle with respect to the Z axis

            tan φ = r / z

where r is the magnitude of the vector in the xy plane

          r = $\sqrt{0.5^{2} + 0.5^{2} }$

          r = 0.7071 m

         φ = tan⁻¹ r / z

         φ = tan⁻¹ (0.7071 / 2)

         φ = 19.5

consequently the total force is

        F_total = 4 F cos 19.5

        F _total =   $4 \frac{k \ q_{1} q_{5} }{r_{15}^{2}}$   cos 19.5

let's calculate

        F_toal = 4 9 10⁹ 1 10⁻⁶ 2 10⁻⁶ /4.5    cos 19.5

        F _total = 15.08 10⁻³ N

b) For this part indicate that the sphere is in equilibrium with the weight

           F_total - W = 0

           W = mg

           F_total = mg

           m = F_total / g

           m = 15.08 10³ / 9.8

           m = 1,539 10⁻³ kg