Two small pith balls, each of mass m = 11.2 g, are suspended from the ceiling of the physics lab by 1.7 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 30.4°, and the charges on the two balls are equal, what is the magnitude of that charge (in µC)? (Please note that the unit of charge is micro-Coulomb here - some browsers might not display the Greek letter mu correctly and show it as an m.)
If the charged ball is suspended be the string which is deflected by the angle α, 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 α = F, ….(1)
y: T•cosα = mg. ….(2)
Divide (1) by (2):
T•sin α/ T•cosα = F/mg,
tan α = F/mg.
Since
q1=q2=q.
r=2•L•sinα,
k=9•10^9 N•m²/C²
F =k•q1•q2/r² = k•q²/(2•L•sinα)².
tan α = F/mg =
= k•q²/(2•L•sinα)² •mg.
q = (2•L•sinα) • sqrt(m•g•tanα/k)=
=(2•1.7•0.506) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =4.5 μC.
To solve this problem, we can use the concept of electrostatic forces and the equilibrium condition.
First, let's define the variables and given information:
- Mass of each pith ball: m = 11.2 g = 0.0112 kg
- Length of each string: L = 1.7 m
- Angle which each string makes with the vertical: θ = 30.4°
Assuming the charges on the two balls are equal, let's denote the magnitude of charge on each pith ball as q.
Now, let's analyze the forces acting on each pith ball:
1. Weight (mg):
The weight of each pith ball acts vertically downward, opposing the upward tension in the string. The weight force is given by the equation:
Weight = mg
2. Tension in the string (T):
The tension in the string acts along the direction of the string and keeps the pith ball in equilibrium. The tension can be split into horizontal and vertical components:
- The horizontal component of tension is balanced by the electrostatic repulsion forces between the charged pith balls.
- The vertical component of tension balances the weight force of the pith ball.
Applying the equilibrium condition, we can find the vertical component of tension:
Vertical component of tension(T_y) = Weight
3. Electrostatic force of repulsion (F_repulsion):
The charged pith balls exert an electromagnetic force of repulsion on each other, which balances the horizontal component of tension.
The electrostatic force of repulsion can be expressed as:
F_repulsion = k * (q^2) / r^2
Here, k is the Coulomb constant, q is the magnitude of charge on each pith ball, and r is the distance between the pith balls.
Since the angle each string makes with the vertical is θ = 30.4°, the angle between the strings is 180° - 2θ.
To find the distance between the pith balls, we can use trigonometry:
r = 2 * L * sin(180° - 2θ)
Now, we can substitute the given values and solve for q:
1. Calculate the vertical component of tension (T_y):
T_y = Weight = mg
2. Calculate the distance between the pith balls (r):
r = 2 * L * sin(180° - 2θ)
3. Calculate the electrostatic force of repulsion (F_repulsion):
F_repulsion = k * (q^2)/ r^2
4. Equate the vertical component of tension with the electrostatic force of repulsion:
T_y = F_repulsion
5. Solve for q:
mg = k * (q^2) / r^2
Rearrange the equation to solve for q:
q = √((m * g * r^2) / k)
Substitute the given values, including the constants:
m = 0.0112 kg
g = 9.8 m/s^2 (acceleration due to gravity)
L = 1.7 m
θ = 30.4°
k = 8.988 × 10^9 N m^2/C^2 (Coulomb constant)
Now, let's calculate q using the formula and the given values.