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•0.5) •sqrt(0.0112•9.8•0.577/9•10^9) =
=4.5•10^-6 C =4.5 μC.
To find the magnitude of the charge on the pith balls, we can use Coulomb's Law and the concept of electrostatic force.
1. First, let's determine the gravitational force acting on each pith ball. We'll use the formula:
Fg = m * g
where m is the mass of each pith ball and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Fg = 11.2 g * (9.8 m/s^2)
Fg = 109.76 g m/s^2
2. Now, let's determine the tension force in the string. The tension force in a string can be broken into vertical and horizontal components. The vertical component balances the gravitational force, and the horizontal component provides the required centripetal force for the circular motion of the pith balls.
Since the pith balls are in equilibrium, the two components must be equal in magnitude:
T * cos(è) = mg (Vertical Component)
T * sin(è) = Fc (Horizontal Component)
where T is the tension force, è is the angle made by each string with the vertical, mg is the gravitational force, and Fc is the centripetal force for the circular motion.
3. To find the centripetal force, we'll first calculate the speed of the pith balls:
The length of the string is given as 1.7 m. When the pith balls are in equilibrium, the vertical component of the tension force provides the gravitational force, and the horizontal component provides the centripetal force.
T * sin(è) = Fc = m * v^2 / r
Since the string length is the radius of the circular motion, we can rewrite it as:
T * sin(è) = m * v^2 / L
Rearranging, we get:
v^2 = (T * sin(è) * L) / m
To find v, we can use the relation:
v = ω * r = ω * L
where ω is the angular velocity, and r is the radius of the circular motion.
v = ω * L
ω * L = √((T * sin(è) * L) / m)
Simplifying, we get:
ω = √((T * sin(è)) / m)
4. Now, let's find the electrostatic force between the pith balls using Coulomb's Law.
The electrostatic force between two charged objects can be given by:
Fe = k * (q1 * q2) / r^2
where Fe is the electrostatic force, k is the Coulomb's constant (approximately 9 × 10^9 N.m^2/C^2), q1 and q2 are the charges on the pith balls, and r is the distance between them.
In our case, the distance between the pith balls is 2L (twice the length of the string), since they are hanging vertically.
Fe = k * (q^2) / (2L)^2
5. Now, let's equate the electrostatic force and the centripetal force:
Fe = * (q^2) / (2L)^2 = m * v^2 / L
Substituting the value of v^2 from step 3:
k * (q^2) / (2L)^2 = m * ((T * sin(è) * L) / m) / L
Simplifying, we get:
q^2 = (T * sin(è) * L^2) / (2 * k)
Taking the square root, we get:
q = √((T * sin(è) * L^2) / (2 * k))
6. Finally, let's substitute the given values and evaluate the expression:
T = mg / cos(è)
L = 1.7 m
k = 9 × 10^9 N.m^2/C^2
m = 11.2 g = 0.0112 kg
è = 30.4°
T = 109.76 g m/s^2 / cos(30.4°)
T = 126.14 N
q = √((126.14 N * sin(30.4°) * (1.7 m)^2) / (2 * 9 × 10^9 N.m^2/C^2))
Evaluating this expression will give you the magnitude of the charge on the pith balls in µC.