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 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•1.8•sin19) •sqrt(0.0166•9.8•tan19/9•10^9) =...
To find the magnitude of the charge on the pith balls, we can use Coulomb's law and the concept of electrostatic forces.
The electrostatic force between two charges is given by Coulomb's law equation: F = k * (|q1| * |q2|) / r^2
In this case, we have two pith balls with equal charges (let's call the magnitude of the charge q). Hence, the force between them can be written as F = k * (q * q) / r^2
Given that the pith balls are not moving, the forces on both balls must balance each other. Let's consider one pith ball and analyze the forces acting on it.
- The weight of the pith ball, mg, acts vertically downwards.
- Tension in the string, T, acts towards the center due to the ball's motion in a horizontal circle.
- Electrostatic force, F, acts along the string due to the repulsion between the charges.
Since the angle that each string makes with the vertical is 19°, the vertical component of the tension force, Tsin(19°), must balance the weight of the pith ball, mg.
Setting up equations of forces:
Tsin(19°) = mg -------- (1)
The horizontal component of the tension force, Tcos(19°), must balance the electrostatic force, F.
Tcos(19°) = F -------- (2)
Since the magnitude of the charges on both pith balls is the same, we can write the electrostatic force equation as:
k * (q * q) / r^2 = F -------- (3)
Now, we'll solve equations (1) and (2) simultaneously to find the tension force, T.
From equation (1):
T = mg / sin(19°)
Substituting this value in equation (2):
(mg / sin(19°)) * cos(19°) = k * (q * q) / r^2
Simplifying the equation:
mg * cos(19°) = (k * q^2 * g) / (r^2 * sin(19°))
Now, let's rearrange the equation to solve for q:
q^2 = (mg * cos(19°) * r^2 * sin(19°)) / (k * g)
Taking the square root of both sides:
q = √((mg * cos(19°) * r^2 * sin(19°)) / (k * g))
Now, plug in the given values:
m = 16.6 g (mass of the pith ball)
r = 1.8 m (length of the string)
è = 19° (angle between the string and the vertical)
k = 8.99 x 10^9 N m^2/C^2 (Coulomb's constant)
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting these values in the equation, we can calculate the magnitude of the charge, q, in µC.