Calculating Distance Travelled Using Electric Potential

In summary, the conversation discusses the forces acting on bodies at maximum velocity, with a focus on electric and frictional forces. It is determined that at maximum velocity, the two horizontal forces are equal to each other and the acceleration is zero. This leads to a calculation error in the book's answer for velocity, which should be 0.67 m/s instead of 2.12 m/s.
  • #1
Jake357
9
2
Homework Statement
Two small bodies have a mass of 0.009 kg each and are tied together with a wire that has a length of 2m. The bodies have the same electric charge of 5*10^-6 C each. The charges are on a horizontal surface. The wire gets cut. Calculate the distance travelled by each body till they stopped and the maximum velocities that they had. Calculate the distance between them at the moment they were having the maximum velocity. The coefficient of friction between the bodies and the surface is 0.1.
The the distance travelled by each body till they stopped must be 5.25 m.
The maximum velocity that they had must be 0.67 m/s.
The distance between them while they are having the maximum velocity must be 5 m.
Relevant Equations
Wf (the work of friction)=μmgd
We (work of the electric potential)=kq^2/d
I only could calculate the distance travelled by each body, by making the difference between the initial and final electric potential work equal to the work of friction done by the 2 bodies.
 
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  • #2
What can you say about the forces on a body when it reaches maximum velocity?
 
  • #3
haruspex said:
What can you say about the forces on a body when it reaches maximum velocity?
The only forces acting on the bodies are: electric, frictional and gravitational. Gravitational is in the y direction, so it doesn't affect the x direction forces, which makes it not useful. The only useful forces are frictional and electric. Their direction is opposite, because the electric force makes them repel each other, but the frictional one wants them to stop so it's in the opposite direction.
And also the electric force is dependent on the distance between them, which means that at a certain distance between them they will have maximum velocity and a different electric force.
 
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  • #4
Jake357 said:
The only forces acting on the bodies are: electric, frictional and gravitational. Gravitational is in the y direction, so it doesn't affect the x direction forces, which makes it not useful. The only useful forces are frictional and electric. Their direction is opposite, because the electric force makes them repel each other, but the frictional one wants them to stop so it's in the opposite direction.
And also the electric force is dependent on the distance between them, which means that at a certain distance between them they will have maximum velocity and a different electric force.
Right, but specifically what can you say about those two horizontal forces when the object is at maximum velocity?
 
  • #5
To @Jake357 : Here is a hint to what @haruspex is trying to convey. A quantity, in this case the speed, is at a maximum at a point where it stops increasing and starts decreasing. What does that imply about the forces acting on the object when there are only two of them?
 
  • #6
haruspex said:
kuruman said:
To @Jake357 : Here is a hint to what @haruspex is trying to convey. A quantity, in this case the speed, is at a maximum at a point where it stops increasing and starts decreasing. What does that imply about the forces acting on the object when there are only two of them?
They are equal to each other.
 
  • #7
haruspex said:
Right, but specifically what can you say about those two horizontal forces when the object is at maximum velocity?
The acceleration is zero, which means that the forces are equal to each other.
 
  • #8
Jake357 said:
The acceleration is zero, which means that the forces are equal to each other.
Right.
 
  • #9
haruspex said:
Right.
kq^2/(2x+l)^2=mgμ
(2x+l)^2=kq^2/(mgμ)=25
2x+l=5
x=5-l/2=1.5 m

kq^2/l=2mv^2/2+kq^2/(2x+l)+2mgμx
mv^2=kq^2/l-kq^2/(2x+l)-2mgμx=0.0405 J
v^2=0.0405/m=4.5
v=2.12 m/s which is wrong. The velocity must be 0.67 m/s.
 
  • #10
Jake357 said:
kq^2/(2x+l)^2=mgμ
(2x+l)^2=kq^2/(mgμ)=25
2x+l=5
x=5-l/2=1.5 m

kq^2/l=2mv^2/2+kq^2/(2x+l)+2mgμx
mv^2=kq^2/l-kq^2/(2x+l)-2mgμx=0.0405 J
v^2=0.0405/m=4.5
v=2.12 m/s which is wrong. The velocity must be 0.67 m/s.
I agree with your working and answer.
Looks like the book answer made a factor of 10 error in calculating ##v^2##.
 

FAQ: Calculating Distance Travelled Using Electric Potential

What is electric potential?

Electric potential, also known as voltage, is the amount of electric potential energy per unit charge at a point in a static electric field. It is measured in volts (V) and represents the work needed to move a positive test charge from a reference point (usually infinity) to a specific point inside the field without producing any acceleration.

How is electric potential related to distance travelled?

Electric potential is related to the work done on a charge and the distance it travels within an electric field. The relationship is given by the equation \( W = q \Delta V \), where \( W \) is the work done, \( q \) is the charge, and \( \Delta V \) is the change in electric potential. By rearranging this formula and knowing the electric field, one can calculate the distance travelled by the charge.

What formula is used to calculate the distance travelled using electric potential?

The formula to calculate distance travelled \( d \) in a uniform electric field \( E \) when a charge \( q \) moves through a potential difference \( \Delta V \) is \( d = \frac{\Delta V}{E} \). This assumes the electric field is uniform and the motion is along the direction of the field.

Can you calculate distance travelled using electric potential in a non-uniform electric field?

In a non-uniform electric field, the calculation is more complex because the electric field strength varies with position. In such cases, one must integrate the electric field over the path of the charge to find the total potential difference and then use this to calculate the distance. The integral form is \( \Delta V = - \int_{a}^{b} \vec{E} \cdot d\vec{l} \).

What are practical applications of calculating distance travelled using electric potential?

Practical applications include designing electric circuits and components, understanding the behavior of charged particles in fields (such as in particle accelerators), and analyzing electric potential in various engineering and physics problems. It is also useful in fields such as electrostatics, electromagnetism, and in the development of various electronic devices.

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