Which two factors affect the electric force between two particles?
1 answer:
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a) Potential energy: 147 m [J]
The gravitational potential energy of an object is given by
where
m is its mass
is the acceleration of gravity
h is the height of the object above the ground
In this problem,
h = 15 m
We call 'm' the mass of the ball, since we don't know it
So, the potential energy of the ball at the top of the hill is
(J)
b) Velocity of the ball at the bottom of the hill: 17.1 m/s
According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:
where
v is the final velocity of the ball
We know from part a) that
U = 147 m
Substituting into the equation above,
And re-arranging for v, we find the velocity:
Answer:
Explanation:
Given the following :
Speed (V) = speed of 2.30×10^7 m/s
Acceleration (a) = 1.70×10^13 m/s^2
Using the right hand rule provided by Lorentz law:
B = F / qvSinΘ
Where B = magnitude of the magnetic field
v = speed of the particle
Θ = 90° (perpendicular to the field)
q = charge of the particle
SinΘ = sin90° = 1
Note F = ma
Therefore,
B = ma / qvSinΘ
Mass of proton = 1.67 × 10^-27
Charge = 1.6 × 10^-19 C
B = [(1.67 × 10^-27) × (1.70 × 10^13)] / (1.6 × 10^-19) × (2.30 × 10^7) × 1
B = 2.839 × 10^-14 / 3.68 × 10^-12
B = 0.7715 × 10^-2
B = 7.72 × 10^-3 T
2) Magnetic field will be in the negative y direction according to the right hand thumb rule.
Since Velocity is in the positive z- direction, acceleration in the positive x - direction, then magnetic field must be in the negative y-direction.
Answer:
a) -1.325 μC
b) 4.17μC
Explanation:
First, you need to know that charge is conserved. So, the adition of the charges, as there is no lost in charge, should always be the same. Also, after the wire is removed, both spheres will have the same charge, trying to find equilibrium. In summary:
We know both q1f and q2f must be positive, because the negative charge at the beginning was the the smaller.
The electrostatic force is equal to:
K is the Coulomb constant, equal to 9*10^9 Nm^2/C^2
Now, we are told that the electrostatic force after the wire is equal to 0.0443 N:
Originally, the force is negative because it was an attraction force, therefore, its direction was opposite to the direction of the repulsive force after the wire:
Solving the quadratic equation:
for this values q_1 wil be:
So as you can see, the negative charge will always be -1.325 μC and the positive 4.17μC