In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.8 m/s. However, this speed is inade
1 answer:
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The difference in electric potential energy between the two points is
where q is the magnitude of the charge and
is the electric potential difference.
But for energy conservation, the difference in electric potential energy
between the two points is equal to the work done to move the charge between A and B:
so we have
and by substituting the numbers of the problem, we find the value of:
where q is the magnitude of the charge and
But for energy conservation, the difference in electric potential energy
so we have
and by substituting the numbers of the problem, we find the value of
As we know that total work done by a force is given by
so it is product of force and displacement along same direction
as we can write it as
so it must be the product of force and displacement in same directions so correct answer must be
<u>B. in the same direction as the displacement vector and the motion.</u>
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. , the amount of charge stored in this capacitor, will stay the same.
The formula relates the electric potential across a capacitor to:
, the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential
by changing the capacitance
of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of
will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of
becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.