Electric Potential-Work Problem

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In summary, the problem involves finding the work required to move a particle of charge Q = +16e from infinity to a point near two fixed particles of charges q1 = +4e and q2 = -2e, given a distance of 1.40 cm and angles of 43 and 60 degrees. The net electric field and potential must be calculated to find the work done, which can be represented as the sum of potentials due to each charge.
  • #1
scoldham
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Homework Statement



Referring to the figure attached, how much work must be done to bring a particle, of charge Q = +16e and initially at rest, along the dashed line from infinity to the indicated point near two fixed particles of charges q1 = +4e and q2 = -2e?

Distance d = 1.40 cm, theta 1 = 43 degrees, and theta 2 = 60 degrees.

Homework Equations



[tex]E = \frac{kQ}{r^2} [/tex]

[tex]v = Ed = \frac{Fd}{q} = \frac{w}{q}[/tex]

The Attempt at a Solution



I've solved for the net E field produced by the 2 still charges. My thought is that I would plug in the magnitude of that value into [tex]Ed = \frac{w}{q}[/tex] along with the charge from the moving particle and that would produce a value for w... But, I do not know what to use for distance... how do I handle the infinity... with a limit?

Help appreciated.
 

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  • Electric Potential-Work Problem.png
    Electric Potential-Work Problem.png
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  • #2
The work must be done to bring the charge Q from infinity (where potential is supposed to be zero) to some point in space is Q times the potential of the electric field in this point (say point A). All you need to do is to find the potential due to [tex]q_1[/tex] and [tex]q_2[/tex] in A. From attached picture you know the distances from [tex]q_1[/tex] and [tex]q_2[/tex] to A, so it's easy to find the potential due to each of charge and then add them to find the total potential. Note also that the work done by the electric field when YOU are moving the charge in it is MINUS the work perfomed by YOU.
 
  • #3
I'm not sure I understand,

Are you saying that

[tex]q_1 E_{net} = v_1[/tex]

and that the same is true for q2 and v2... so,

[tex]v_1+v_2 =\Delta V[/tex]

Therefore

[tex]w = Q \Delta V[/tex]
 
  • #4
I'm saying that the work done is
[tex]W = - Q \int_{\infty}^{A} \mathbf{E} \mathbf{dl} = Q \left[ \varphi(A) - \varphi(\infty) \right][/tex]
If you set [tex]\varphi ({\infty) = 0[/tex] (remember that potential is only defined up to an additive constant) then
[tex]W = Q \varphi (A)[/tex]
where
[tex]\varphi (A) = \varphi_{q1} + \varphi_{q2}[/tex]
that is the sum of two potentials due to each of the charge.
 
Last edited:
  • #5


I would approach this problem by first breaking it down into smaller parts. I would start by calculating the electric field at the point where the particle is initially at rest. This can be done by using the equation E = \frac{kQ}{r^2}, where Q is the charge of the particle and r is the distance from the particle to the fixed charges.

Next, I would calculate the electric potential at the same point. This can be done by using the equation V = \frac{kQ}{r}, where Q is the charge of the particle and r is the distance from the particle to the fixed charges.

Once I have these values, I would use the fact that the work done on a particle is equal to the change in its kinetic energy, which can be represented by the equation W = \Delta KE = \frac{1}{2}mv^2. Since the particle is initially at rest, its initial kinetic energy is zero. Therefore, the work done on the particle is equal to its final kinetic energy, which can be calculated using the electric potential and the charge of the particle.

Since the particle is moving along a curved path, we can use the equation v = Ed to calculate the final velocity of the particle. This equation represents the relationship between the electric field, the distance traveled, and the final velocity of the particle. Once we have the final velocity, we can plug it into the equation W = \Delta KE = \frac{1}{2}mv^2 to calculate the work done on the particle.

In terms of handling the infinity in the problem, we can use the concept of a limit. As the particle gets closer and closer to the fixed charges, the distance between them becomes smaller and smaller, and the electric field and potential approach a finite value. Therefore, we can use the value of the electric field and potential at a very large distance as an approximation for the values at infinity.

In summary, to calculate the work done on the particle, we need to calculate the electric field and potential at the point where the particle is initially at rest, use the electric field to calculate the final velocity of the particle, and then use the final velocity to calculate the work done on the particle.
 

What is electric potential?

Electric potential is the amount of work needed to move a unit positive charge from one point to another in an electric field. It is measured in volts (V).

How is electric potential related to electric field?

The electric potential at a point is directly proportional to the electric field at that point. The electric field is the gradient of the electric potential.

How can electric potential be calculated?

The electric potential at a point can be calculated by dividing the work done in moving a unit positive charge from infinity to that point by the amount of charge. This can be represented by the equation V = W/Q.

What is a common application of electric potential?

Electric potential is commonly used in the design of electronic circuits and devices. It helps engineers determine the amount of energy required to power a device and how it will behave in different electric fields.

How can electric potential be changed?

Electric potential can be changed by altering the distance between two charges, the magnitude of the charges, or the medium between the charges. It can also be changed by moving the charges themselves.

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