The thick lines represent hollow plastic tubes through which a charged head coul

Question

The thick lines represent hollow plastic tubes through which a charged head could alide frictionleasly from point 1. to point N or vice versa. 1. A positively charged head is put into the straight tube at L. Does it naturally move to point N, or must it be pushed through the tube to reach point N? Explain. 2. Two identical heads (same charge, same mass) are released from real into the straight and curvy tube respectively. at point L. Which bead (if either) reaches point with more speed? Explain. 3. A student points out, correctly, that the head in the bead in the straight tube has a higher acceleration, on average, than the bead in the curve tube. Reconcile this observation with your answer to the previous question.

Explanation / Answer

Hi,

In this case we have to remember certain thing about electric potential and some of current and resistance.

1. The notion of electric potential can be understood if we imagine a diference of potential as a difference in height while at the same time we imagine that the positive charges are like masses.

If we have a mass that is a certain level above the ground (a region with less height), this will fall freely and will gain speed. The same happens if we put a positive charge in a region where there is a difference in electric potential. The charge will move towards a region with less potential from a region of higher potential and in the process it will gain speed.

Therefore, a positively charged bead will move on its own from L to N.

2. The nature of the electric potential makes it a conservative field, therefore, any work done by it will be independent of the path the object chooses. Of course, this is true as long as there are no non conservative forces (such as the friction),

In our case, the speed of any bead will be determine thanks to the conservation of energy:

We consider two points, L and N.

At L the only energy (considering that we are not in presence of any gravitational field) will be the electric potential energy which is:

EL = qV

On the other hand, at N, the only energy will be the kinetic one, so:

EN = (1/2) mv2

Both energies should be equal as the walls of the tubes are frictioneless, so:

v2 = 2qV/m ; as we can see the final speed is proportional to the mass, the difference of potential an the electric charge.

As all those parameters are the same to both beads, we have to say that they reach N with the same speed.

3. To work with this new information, we can use some equations from kinematics.

v = vo + at ; where vo is the initial speed of any bead (which is cero, as both are at rest at the beginning of their movement), a is the averga acceleration and t is variation in time.

A sub C will indicate that we are mentioning parameters of the curvy tube, while a sub S will mean that we are talking about parameters of the straight tube.

vc = ac tc  ; vs = as ts ::::::: (vc/vs ) = [ ac tc / as ts] ; as vc = vs :::::::: ac tc = as ts

Then, if as > ac this means that tc > ts ; which is logic as the bead in the curvy tube will take more time to reach N.

4. Electric circuits.

In a bulb there is a resistor that transforms the electric energy that the battery supplies into light and heat. The bright of a bulb is proportional to the power its resistance is able to resistor from the battery.

Said power can be calculated as:

P = I2 R ; where I is the current through the resistor and R is the resistance of said resistor.

If we have bulbs as the ones we see in the image, then the current will have to be calculated using an equivalent resistance, which in this case as the bulbs are in series, is equal to the the sum of all the resistances:

Req = R1 + R2 + R3 so I = emf/Req and P = (emf/Req)2 R

As we can see, the value of current is common to any bulb. The difference would be determined if any of them had a different resistance, but as they are all the same, all of them have the same bright.

I hope it helps.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.