Pith Balls
Two small pith balls, each of mass m = 17.7 g, are suspended from the ceiling of the physics lab by 0.8 m long fine strings and are not moving. If the angle which each string makes with the vertical is è = 42.6°, 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•0.8•sin42.6) •sqrt(0.0177•9.8•tan42.6/9•10^9) =...
sqrt((0.0177(2*0.8*sin42.6))(tan42.6)/9*10^9)
sqrt((0.0177(2*0.8*sin42.6))(tan42.6)/9*10^9)=1.5*10^-6
From what height can a 1-kg falling object cause fracture of the skull? Assume that the object is hard, that the area of contact with the skull is 1 cm2, and that the duration of impact is 10−3 sec.
To find the magnitude of the charge on the pith balls, we can use Coulomb's law which states that the electric force between two charged objects is given by:
F = k * |q1 * q2| / r^2
where:
- F is the magnitude of the electric force,
- k is Coulomb's constant (k = 8.99 x 10^9 N*m^2/C^2),
- q1 and q2 are the charges on the pith balls, and
- r is the distance between the centers of the pith balls.
In this case, both pith balls have the same charge and are in equilibrium. This means that the weight of each pith ball is balanced by the electrostatic repulsion between them.
First, let's find the tension in each string:
T = m * g
where:
- T is the tension in the string,
- m is the mass of each pith ball, and
- g is the acceleration due to gravity (approximately 9.8 m/s^2).
T = (0.0177 kg) * (9.8 m/s^2) = 0.17346 N
Now, because the strings are at an angle with the vertical, the vertical component of the electrostatic force must balance the weight of the pith ball. Therefore, we can write:
T * cos(θ) = F
where:
- θ is the angle each string makes with the vertical.
Rearranging the equation, we get:
F = T * cos(θ)
Now, since both pith balls have the same charge, the magnitudes of the electrostatic forces acting on them will also be the same. Therefore, we can equate the magnitudes of the forces:
F = F
T * cos(θ) = F
We can substitute the expression for F from Coulomb's law:
T * cos(θ) = (k * |q|^2) / r^2
But we need to find |q|, the magnitude of the charge. So, let's solve for |q|:
|q|^2 = (T * cos(θ) * r^2) / k
|q| = sqrt((T * cos(θ) * r^2) / k)
Plugging in the given values:
|q| = sqrt((0.17346 N * cos(42.6°) * (0.8 m)^2) / (8.99 x 10^9 N*m^2/C^2))
Calculating this expression will give us the magnitude of the charge on the pith balls in Coulombs (C). To convert it to micro-Coulombs (µC), we divide the result by 10^-6.
So, the final step is to divide the magnitude of the charge by 10^-6:
|q| = (calculated value) / 10^-6
Evaluating this expression will give you the magnitude of the charge on the pith balls in micro-Coulombs (µC).