What is the correct order from first stage of development to the last
1 answer:
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The answer is: 40 J.
To solve this problem you must apply the proccedure shown below:
1. You have the following formula that is used to calculate the kinetic energy:
Where m is the mass of the objetct and V is the speed.
2. Substitute values into the formula:
3. Therefore, the result is:
(a) the electric field generated by a single point charge is radial, and its direction is determined by the sign of the source charge. In particular, when the source charge is positive as in this example, the field lines are directed away from the charge (see attached figure).
(b) The electric field strength at distance r from the source charge is given by
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
The force experienced by the test charge in this electric field is equal to
where
is our test charge. Substituting, we find that the force is equal to
(c) The electric potential at a distance r from the source (positive) charge is given by
and the electric potential energy of a test charge q at distance r from the source charge is
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.
(b) The electric field strength at distance r from the source charge is given by
where ke is the Coulomb's constant and Q is the source charge. In our problem, the distance is r=2.00 cm=0.02 m, so the electric field at that point is
The force experienced by the test charge in this electric field is equal to
where
(c) The electric potential at a distance r from the source (positive) charge is given by
and the electric potential energy of a test charge q at distance r from the source charge is
We can see that when the test charge q is moved closer to the source Q, the distance r decreases, and so the potential energy U of the charge increases. So, the change in potential energy is positive.