The figure below shows two identical small, charged spheres. One of mass 3.7 g is hanging by an insulating thread of length 20.0 cm. The other is held in place and has charge
q1 = −5.3 µC.
The thread makes an angle of 18° with the vertical, resulting in the spheres being aligned horizontally, a distance r apart. Determine the charge q2 on the hanging sphere.

? nC

Also, the horizontal component of the tension,T" equals the coulomb force of repulsion, F between the two charges q₁ and q₂ separated by a distance r.

So, T" = F

Tsin18° = kq₁q₂/r² (2)

Diving equation (2) by (1), we have

Tsin18° ÷ Tcos18° = kq₁q₂/r² ÷ mg

tan18° = kq₁q₂/(r²mg)

The q₂ charge on the hanging sphere

Making q₂ subject of the formula, we have

q₂ = r²mgtan18°/kq₁

From the diagram, sin18° = r/20.0 cm

r = 20sin18° cm = 0.20sin18° m = 0.062 m

Given that

q₁ = charge on fixed sphere = -5.3 μC = -5.3 × 10⁻⁶ C,
m = mass of hanging sphere = 3.7 g = 3.7 × 10⁻³ kg,
k = coulomb constant = 9 × 10⁹ Nm²/C², and
g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

q₂ = r²mgtan18°/kq₁

q₂ = (0.062 m)² × 3.7 × 10⁻³ kg × 9.8 m/s² × tan18°/(9 × 10⁹ Nm²/C² × 5.3 × 10⁻⁶ C)

q₂ = 0.003844 m² × 3.7 × 10⁻³ kg × 9.8 m/s² × tan18°/(9 × 10⁹ Nm²/C² × 5.3 × 10⁻⁶ C)

q₂ = 0.13938344 × 10⁻³ m³kg/s² × 0.3249/(9 × 10⁹ Nm²/C² × 5.3 × 10⁻⁶ C)

q₂ = 0.0453 × 10⁻³ m³kg/s²/(47.7 × 10³ Nm²/C)

q₂ = 0.00095 × 10⁻⁶ C

q₂ = 9.5 × 10⁻¹⁰ C

q₂ = 0.95 × 10⁻⁹ C

q₂ = 0.95 nC

Since the force acting on both charges is repulsive, q₂ = -0.95 nC

So, the charge q₂ on the hanging sphere is -0.95 nC

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