A rectangular tank that is 4 meters long, 3 meters wide and 6 meters deep is filled with a rubbing alcohol that has density 786
kilograms per cubic meter. In each part below, assume that the tank is initially full, and that gravity is 9.8 meters per second squared. Your answers must include the correct units.
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?
(a) How much work is done pumping all of the liquid out over the top of the tank?
(b) How much work is done pumping all of the liquid out of a spout 2 meters above the top of the tank?
(c) How much work is done pumping two-thirds of the liquid out over the top of the tank?
(d) How much work is done pumping two-thirds of the liquid out of a spout 2 meters above the top of the tank?
1 answer:
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Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
Now At xp we have:
Which is a second order equation, using the quadratic formula to solve for xp would give us:
or
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.
Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
Now At xp we have:
Which is a second order equation, using the quadratic formula to solve for xp would give us:
or
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.
Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.