Two small pith balls, each of mass m = 16.6 g, are suspended from the ceiling of the physics lab by 1.8 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 19°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)?

If the charged ball is suspended be the string which is deflected by the angle á, the forces acting on it are: mg (downwards), tension T (along the string - to the pivot point), and F (electric force –along the line connecting the charges).

Projections on the horizontal and vertical axes are:
x: T•sin á = F, ….(1)
y: T•cosá = mg. ….(2)
Divide (1) by (2):
T•sin á/ T•cosá = F/mg,
tan á = F/mg.

Since
q1=q2=q.
r=2•L•siná,
k=9•10^9 N•m²/C²
F =k•q1•q2/r² = k•q²/(2•L•siná)².

tan á = F/mg =
= k•q²/(2•L•siná)² •mg.
q = (2•L•siná) • sqrt(m•g•taná/k)=
=(2•1.7•0.506) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =8182918 ìC.

To find the magnitude of the charge on the pith balls, you need to use the principle of electrostatic equilibrium. In this situation, the gravitational force acting on each pith ball is balanced by the electrostatic force due to the charges on the balls.

Here's how you can find the magnitude of the charge on the pith balls:

1. Start by calculating the weight (mg) of each pith ball. The mass of each ball is given as 16.6 g, so the weight of each ball is:

weight = mass × acceleration due to gravity
= 16.6 g × 9.8 m/s²
= 0.0166 kg × 9.8 m/s²

2. Next, calculate the tension in each string. The tension in each string can be represented as the sum of the vertical component of the tension and the horizontal component of the tension. The vertical component of the tension balances the weight of each ball, while the horizontal component is responsible for balancing the electrostatic forces.

Tension in string = vertical component of tension + horizontal component of tension

The vertical component of tension is given by:
vertical component of tension = weight / cos(θ) [θ is the angle made by the string with the vertical]

Substitute the known values:
vertical component of tension = weight / cos(19°)

The horizontal component of tension is given by:
horizontal component of tension = electrostatic force / sin(θ)

Since the two balls have equal charges, the electrostatic force on each ball is equal. Therefore, you can write:
horizontal component of tension = electrostatic force × 2 / sin(19°) [multiplying by 2 to account for both balls]

3. Set the total tension in the string equal to the sum of the vertical and horizontal components:

vertical component of tension + horizontal component of tension = tension in string

Substitute the known values:
weight / cos(19°) + electrostatic force × 2 / sin(19°) = tension in string

4. Rearrange the equation to solve for the electrostatic force:

electrostatic force = (tension in string - weight / cos(19°)) × sin(19°) / 2

5. Finally, to find the magnitude of the charge on the pith balls, use Coulomb's law, which states that the electrostatic force between two charges is given by:

electrostatic force = (k × q₁ × q₂) / r²

where k is the electrostatic constant (9 × 10⁹ N m²/C²), q₁ and q₂ are the magnitudes of the charges, and r is the distance between the charges.

In this scenario, the charges on the pith balls are the same magnitude, so q₁ = q₂ = q. Additionally, the distance between the balls is equal to the length of the string, which is given as 1.8 m.

Substituting these values into Coulomb's law and equating it to the electrostatic force from step 4, solve for q:

(k × q²) / r² = (tension in string - weight / cos(19°)) × sin(19°) / 2

Rearrange and solve for q:

q = √[((tension in string - weight / cos(19°)) × sin(19°) × r²) / (k)]

So, plug in the known values:

q = √[((tension in string - weight / cos(19°)) × sin(19°) × (1.8 m)²) / (9 × 10⁹ N m²/C²)]

Calculate the tension in the string using the vertical component of tension:

tension in string = weight / cos(19°)

Finally, substitute the calculated values back into the equation for q to get the magnitude of the charge on the pith balls in µC (microcoulombs).