An electric field of 1139 V/m is applied to a section of silver of uniform cross section. Find the resulting current density if
1 answer:
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Here we can say that rate of flow must be constant
so here we will have
now we know that
now from above equation
so velocity will reduce by factor 0.14
A) The work done by the electric field is zero
B) The work done by the electric field is
C) The work done by the electric field is
Explanation:
A)
The electric field applies a force on the charged particle: the direction of the force is the same as that of the electric field (for a positive charge).
The work done by a force is given by the equation
where
F is the magnitude of the force
d is the displacement of the particle
is the angle between the direction of the force and the direction of the displacement
In this problem, we have:
- The force is directed vertically upward (because the field is directed vertically upward)
- The charge moves to the right, so its displacement is to the right
This means that force and displacement are perpendicular to each other, so
and : therefore, the work done on the charge by the electric field is zero.
B)
In this case, the charge move upward (same direction as the electric field), so
and
Therefore, the work done by the electric force is
and we have:
is the magnitude of the electric force. Since
is the magnitude of the electric field
is the charge
The electric force is
The displacement of the particle is
d = 0.660 m
Therefore, the work done is
C)
In this case, the angle between the direction of the field (upward) and the displacement (45.0° downward from the horizontal) is
Moreover, we have:
(electric force calculated in part b)
While the displacement of the charge is
d = 2.50 m
Therefore, we can now calculate the work done by the electric force:
And the work is negative because the electric force is opposite direction to the displacement of the charge.
Learn more about work and electric force:
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