A positive test charge q is released from rest at distance r away from a charge of Q and a distance 2r away from a charge of 2Q.
1 answer:
Answer: Option (b) is the correct answer.
Explanation:
It is given that a positive test charge q is released from rest at a distance r away from a charge of +Q and a distance 2r which is away from a charge of +2Q.
Then test charge to the right immediately after being released.
Therefore, the net force will be as follows.
F =
=
=
F = > 0
Thus, we can conclude that the test charge move to the right immediately after being released.
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There is a missing data in the text of the problem (found on internet):
"with force constant<span> k=</span>450N/<span>m"
a) the maximum speed of the glider
The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x
When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>
<span>
Since the mechanical energy must be conserved, we can write
</span>
<span>from which we find the maximum speed
</span>
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m
We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span>
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>
<span>And since
</span>
<span>we find the kinetic energy when the glider is at this position:
</span>
<span>And then we can find the corresponding velocity:
</span>
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
where
is the angular frequency, and A is the amplitude.
The angular frequency is:
and so the maximum acceleration is
d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m
For a simple harmonic motion, the acceleration is given by
</span>
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span>
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>
we have already calculated it at point b), and it is given by
</span>
"with force constant<span> k=</span>450N/<span>m"
a) the maximum speed of the glider
The total mechanical energy of the mass-spring system is constant, and it is given by the sum of the potential and kinetic energy:
</span>
<span>where
k is the spring constant
x is the displacement of the glider with respect to the spring equilibrium position
m is the glider mass
v is the speed of the glider at position x
When the glider crosses the equilibrium position, x=0 and the potential energy is zero, so the mechanical energy is just kinetic energy and the speed of the glider is maximum:
</span>
<span>Vice-versa, when the glider is at maximum displacement (x=A, where A is the amplitude of the motion), its speed is zero (v=0), therefore the kinetic energy is zero and the mechanical energy is just potential energy:
</span>
<span>
Since the mechanical energy must be conserved, we can write
</span>
<span>from which we find the maximum speed
</span>
<span>
b) </span><span> the </span>speed<span> of the </span>glider<span> when it is at x= -0.015</span><span>m
We can still use the conservation of energy to solve this part.
The total mechanical energy is:
</span>
<span>
At x=-0.015 m, there are both potential and kinetic energy. The potential energy is
</span>
<span>And since
</span>
<span>we find the kinetic energy when the glider is at this position:
</span>
<span>And then we can find the corresponding velocity:
</span>
<span>
c) </span><span>the magnitude of the maximum acceleration of the glider;
</span>
For a simple harmonic motion, the magnitude of the maximum acceleration is given by
where
The angular frequency is:
and so the maximum acceleration is
d) <span>the </span>acceleration<span> of the </span>glider<span> at x= -0.015</span><span>m
For a simple harmonic motion, the acceleration is given by
</span>
<span>where x(t) is the position of the mass-spring system. If we substitute x(t)=-0.015 m, we find
</span>
<span>
e) </span><span>the total mechanical energy of the glider at any point in its motion. </span><span>
we have already calculated it at point b), and it is given by
</span>