We have that the electric field at the center of the metal ball due only to the charges on the surface of the metal ball is
From the question we are told that
A solid metal ball of radius 1.5 cm
bearing a charge of -15 nC is located near a hollow plastic ball of radius 1.9 cm bearing
uniformly distributed charge of -7 nC
The distance between the centers of the balls is 9 cm
Generally the equation for the electric field is mathematically given as
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Answer:
w=3.05 rad/s or 29.88rpm
Explanation:
k = coefficient of friction = 0.3900
R = radius of the cylinder = 2.7m
V = linear speed of rotation of the cylinder
w = angular speed = V/R or to rewrite V = w*R
N = normal force to cylinder
N=
These must be balanced (the net force on the people will be 0) so set them equal to each other.
There are 2*pi radians in 1 revolution so:
So you need about 30 RPM to keep people from falling out the bottom
Answer:
x = 0.775m
Explanation:
Conceptual analysis
In the attached figure we see the locations of the charges. We place the charge q₃ at a distance x from the origin. The forces F₂₃ and F₁₃ are attractive forces because the charges have an opposite sign, and these forces must be equal so that the net force on the charge q₃ is zero.
We apply Coulomb's law to calculate the electrical forces on q₃:
(Electric force of q₂ over q₃)
(Electric force of q₁ over q₃)
Known data
q₁ = 15 μC = 15*10⁻⁶ C
q₂ = 6 μC = 6*10⁻⁶ C
Problem development
F₂₃ = F₁₃
(We cancel k and q₃)
q₂(2-x)² = q₁x²
6×10⁻⁶(2-x)² = 15×10⁻⁶(x)² (We cancel 10⁻⁶)
6(2-x)² = 15(x)²
6(4-4x+x²) = 15x²
24 - 24x + 6x² = 15x²
9x² + 24x - 24 = 0
The solution of the quadratic equation is:
x₁ = 0.775m
x₂ = -3.44m
x₁ meets the conditions for the forces to cancel in q₃
x₂ does not meet the conditions because the forces would remain in the same direction and would not cancel
The negative charge q₃ must be placed on x = 0.775 so that the net force is equal to zero.
