Two identical spheres are each attached to silk threads of length l= 0.500 m and hung from a common point. each sphere has mass m= 8.00 g. the radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. one sphere is given positive charge q1, and the other a different positive charge q2 ( q2 < q1 ); this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle θ= 20.0 degrees with the vertical.

a small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. each thread now makes an angle of 30.0 degrees with the vertical. determine the original charges q1 and q2. (hint: the total charge on the pair of spheres is conserved.) q1= coulombs, q2 = coulombs

q1 = 1.68 10⁻⁶ C    and    q2 = 0.56 10⁻⁶ C

Explanation:

Let's write Newton's second law for the two spheres, let's start by looking for the stress components with trigonometry

        sin θ = Tx / T

        Tx = T sin θ

        Cos θ = Ty / T

        Ty = cos θ

Case 1 q1 = q2

        Fe -Tx = 0

        Ty -W = 0

Case 2 the charge (q) is the same for the two spheres,

        Fe-Tx = 0

        Ty -W = 0

        T cos θ = mg

        T = mg / cos θ

        k q q / r² - T sin 30 = 0

        k q2 / r² - mg / cos 30 sin 30 = 0

        k q2 / r² = mg tan 30

Let's look for the separation of the spheres, using trigonometry

        sin 30 = a / L

        a = L sin 30

        a = 0.500 sin 30

        a = 0.25

This is the length from a sphere to the midpoint of the triangle, whereby the distance between the two spheres is double

       r = 2 a = 2 0.25

       r = 0.500 m

Let's calculate the charge

       k q² / r² = mg tan 30

       q2 = mg r² /k  tan 30

       q2 = 8 10⁻³  9.8 0.5² /8.99 10⁹ tan 30

       q2 = 1.26 10⁻¹² C2

Now let's work in case 1 q1≠q2

We have the same equations, but the angle T = 20º

       a = L sin 20

       r = 2 a

       k q1 q2 / r2 = mg tan 20

Let's calculate the charges

       a = 0.500 sin 20

       a = 0.171 m

       r = 0.342 m

       q1q2 = mg r² / k tan 20

       q1q2 = 8 10⁻³  9.8 0.342² / 8.99 10⁹ tan 20

       q1q2 = 0.371 10⁻¹² C2

Let's write the system of equations

      q2 = 1.26 10⁻¹²              (1)

      q1 q2 = 0.371 10⁻¹²

      q1 + q2 = Q

When analyzing this system of equations, we see that the load (q) of the spheres should be half of the total load

       q = Q / 2

With (1)

       Q² / 4 = 1.26 10⁻¹²

       Q = √ 4 1.26 10⁻¹²

       Q = 2.24 10⁻⁶ C

We solve the other two equations

       (Q -q2) q2 = 0.371 10-12

       Q q2 - q22 = 0.371 10-12

       q2 2 -2.24 10-6 q2 + 0.371 10-12

We solve the second degree equation

       q2 = {2.24 10⁻⁶ + - √ [(2.24 10⁻⁶) 2 - 4 1  0.371 10⁻¹²]} / 2

       q2 = (2.24 10⁻⁶ + - √ 3,534 10⁻¹²) / 2

       q2 = 2.24 10⁻⁶ + - 1.88 10⁻⁶) 2

       q2 ’= 2.16 10⁻⁶ C

       q2 ’’ = 0.56 10⁻⁶ C

We calculate q1

       q1 + q2 = 2.24 10⁻⁶

       q1 ’= 2.24 10⁻⁶ - q2’

       q1 ’’ = 2.24 10-⁻⁶ -q2 ’’

       q1 ’= 0.08 10⁻⁶ C

       q1 ’’ = 1.68 10⁻⁶ C

As they indicate that q2 <q1 the correct solution is

       q1 = 1.68 10⁻⁶ C

       q2 = 0.56 10⁻⁶ C

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