The Electric Field "in front of" and "in back of" Two Point Charges

G.R.Dixon, September 19, 2008

The general formula for the electric field of point charge q is

(1)

In this formula **e**_{r} is a unit vector from the charges *retarded* position to the field evaluation point. And r is the scalar distance from the retarded position to the field evaluation point.

In the present case we consider two point charges, q_{1} and q_{2}, whose common acceleration is constant, and whose positions are

, (2a)

. (2b)

(t is limited to values such that a|t|<c. Prior to t=0 the charges are moving in the negative x direction. At t=0 they come to rest at x=0 and x=r respectively. At all t>0 they move in the positive x direction.)

Since the time derivatives of **e**_{r} are zero in Eq. 1, we may simplify that formula to

. (3)

Or, defining E_{1} to be q_{1}s field at the momentary position of q_{2} and E_{2} to be q_{2}s field at the momentary position of q_{1}, we have

, (4a)

. (4b)

Fig. 1 depicts the situation at some t<0. Note that r_{1}<r and r_{2}>r. Thus

. (5)

Figure 1

r_{1} and r_{2} at some t<0

The formulas for r_{1}, dr_{1}/dt, r_{2} and dr_{2}/dt are

, (6a)

, (6b)

, (6c)

. (6d)

(These formulas are derived in Appendix A at the end of the article.) Plugging these values into Eqs. 4a and 4b produces E_{1} and E_{2}.

Now when **v** is *constant*, the formulas for E_{1} and E_{2} are considerably simpler:

, (7a)

. (7b)

An interesting question is whether the values of E_{1} and E_{2} in the accelerated case equal those in the constant-velocity case. The question is readily answered by plotting the values for the two cases over a range of velocity components, say

. (8)

(A Visual Basic program that does this is provided in Appendix B.) Fig. 2 plots the accelerated values for E_{1} and E_{2} and the constant-velocity values. (The top curve is an overlay of the fields for q_{1}; the bottom curve is an overlay of the fields for q_{2}.) Note the perfect overlays. Evidently the "forward" and "backward" looking electric fields of a charge moving with constant velocity are the same as those of a charge subjected to a constant acceleration.

**Figure 2**

E_{1} and E_{2}, constant v and constant a cases

***Appendix A***

In this appendix the formula for E_{1} is derived for any time t<0. (q_{1} and q_{2} are moving in the negative x direction, and are being accelerated in the positive x direction.) The cases for q_{2}, and for t>0, are similarly derived.

Fig. A1 depicts the retarded position, etc., for q_{1}.

Figure A1

r_{1}, etc.

The retarded time is defined to be the present time minus a positive Dt.

. (A1)

The retarded velocity component is the constant acceleration times the retarded time.

. (A2)

Dx is the (positive) distance between the present position and the retarded position.

, (A3)

. (A4)

But

(see Fig. A1), (A5)

And by definition,

. (A6)

Subtracting A6 from A5:

. (A7)

And substituting from A4 produces

. (A8)

Solving A8 for Dt_{1} produces

. (A9)

And, since

, (A10)

differentiation of A9 produces

. (A11)

Finally,

(A12)

and

. (A13)

These can now be plugged into the formula for E_{1}:

. (A14)

***Appendix B***

The Visual Basic program used to compute the fields follows.

Private Sub cmdConAccelE_Click()

Const N As Long = 1000

Const c As Double = 300000000# 'speed of light

Const vinit As Double = -0.99 * c

Const vfinal As Double = -0

Const dv As Double = (vfinal - vinit) / N

Const r As Double = 1 'separation of the charges

Const a As Double = 1000000 'constant acceleration

Const q As Double = 1 'magnitude of charges

Const eps0 As Double = 0.00000000000885

Const pi As Double = 3.141592654

Dim index As Long

Dim v(2 * N) As Double

Dim E1(2 * N), E2(2 * N), E(2 * N), E1p(2 * N), E2p(2 * N) As Double

Dim dt1, dt2, ddt1dt, ddt2dt As Double 'retarded time

Dim rp1, rp2, drp1dt, drp2dt As Double 'distance from retarded position to field eval pt

Open "c:/WINMCADC/Physics/TwoQ.PRN" For Output As #1

For index = 0 To N - 1

'For t<0

Debug.Print index

'and for each velocity considered,

v(index) = vinit + index * dv

'compute the constant-velocity fields.

E(index) = q / (4 * pi * eps0 * r ^ 2) * (1 - v(index) ^ 2 / c ^ 2)

E1p(index) = E(index)

E2p(index) = -E(index)

'Then compute the constant acceleration fields.

dt1 = -((c - v(index)) / a) + Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a)

ddt1dt = 1 - c / (a * Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a))

rp1 = c * dt1

drp1dt = c * ddt1dt

E1(index) = q / (4 * pi * eps0 * rp1 ^ 2) * (1 - 2 / c * drp1dt)

dt2 = (c + v(index)) / a - Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a)

ddt2dt = 1 - c / (a * Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a))

rp2 = c * dt2

drp2dt = c * ddt2dt

E2(index) = -q / (4 * pi * eps0 * rp2 ^ 2) * (1 - 2 / c * drp2dt)

Write #1, v(index), E1(index), E1p(index), E2(index), E2p(index)

Next index

For index = N To 2 * N - 1

'Do the same thing for times t>0.

Debug.Print index

v(index) = vfinal + (index - N) * dv

E(index) = q / (4 * pi * eps0 * r ^ 2) * (1 - v(index) ^ 2 / c ^ 2)

E1p(index) = E(index)

E2p(index) = -E(index)

dt1 = -((c - v(index)) / a) + Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a)

ddt1dt = 1 - c / (a * Sqr(((c - v(index)) / a) ^ 2 + 2 * r / a))

rp1 = c * dt1

drp1dt = c * ddt1dt

E1(index) = q / (4 * pi * eps0 * rp1 ^ 2) * (1 - 2 / c * drp1dt)

dt2 = (c + v(index)) / a - Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a)

ddt2dt = 1 - c / (a * Sqr(((c + v(index)) / a) ^ 2 - 2 * r / a))

rp2 = c * dt2

drp2dt = c * ddt2dt

E2(index) = -q / (4 * pi * eps0 * rp2 ^ 2) * (1 - 2 / c * drp2dt)

'Output all of the field values for plotting purposes.

Write #1, v(index), E1(index), E1p(index), E2(index), E2p(index)

Next index

Close

MsgBox ("Ready for plotting")

Stop

End Sub