Electric Potential-Work Problem

[scoldham]

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

E = \frac{kQ}{r^2}

v = Ed = \frac{Fd}{q} = \frac{w}{q}

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 Ed = \frac{w}{q} 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.

 

[physicsworks]

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 q_1 and q_2 in A. From attached picture you know the distances from q_1 and q_2 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.

 

[scoldham]

I'm not sure I understand,

Are you saying that

q_1 E_{net} = v_1

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

v_1+v_2 =\Delta V

Therefore

w = Q \Delta V

 

[physicsworks]

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